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•1 ELECTRONIC ENGINEERING 



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WILEY 
TOP PAN 



C. L. Alley and K . W. Atwood 




V 



Wiley International Edition 



Electronic 

Engineering 

SECOND EDITION 

Charles L. Alley Sl Kenneth W. Atwood 

UNIVERSITY OF UTAH, ELECTRICAL ENGINEERING DEPARTMENT 



JOHN WILEY & SONS, INC. NEW YORK LONDON SYDNEY 
TOPPAN COMPANY, LTD., Tokyo, Japan 



Copyright © 1962, 1966 by John Wiley & Sons, Inc. 

All Rights Reserved 

This book or any part thereof must not 
be reproduced in any form without the 
written permission of the publisher. 

Wiley International Edition 

This book is not to be sold outside the country 
to which it is consigned by the publisher. 

Library of Congress Catalog Card Number: 66-16128 

Printed in Japan 
By TOPPAN PRINTING COMPANY. LTD. 



To Mildred and Ruth 

whose patience and understanding 

made this book possible. 



Prefc 



ace 



The rapidly changing field of electronics requires a continual re-evaluation 
of the content of an electronics textbook. Also, improved methods of 
presenting familiar subjects are constantly being sought by a resourceful 
teacher. For these reasons, we felt that a new edition would be very 
beneficial to those teachers who strive to keep their material up to 
date. Also, the practicing engineer requires reference material which is 
current. 

Great assistance in planning this second edition was given by the 
publisher, who mailed questionnaires to a large number of users of the 
first edition. Criticism of the first edition and recommendations for 
the second edition were solicited and generously given. We are deeply 
indebted to these many contributors for their response and excellent 
suggestions. Although it was impossible to incorporate all of these 
suggestions into a single volume, the second edition does incorporate a 
great number of these recommendations. In addition, the publisher 
obtained outstanding reviews of the revised material, which resulted in a 
considerable improvement of the text. To all of those who have given 
their time to this cause, we express our heartfelt thanks. 

The major changes in the second edition include the following ones. 

1. IEEE standard notation has been adopted. 

2. Some of the newer devices such as field-effect transistors and uni- 
junction transistors have been included. 

3. A new chapter on cascaded amplifiers and noise has been included. 

4. Emphasis on the semiconductor devices has been increased. 

5. Chapter 16 on pulse applications has been completely rewritten and 
major changes have been made in most of the remaining chapters in 
order to achieve the aforementioned goals. 



viii Preface 

6. A large number of problems have been rewritten and many new 
comprehensive problems have been added. 

7. By deleting some of the less important material and improving the 
efficiency of presentation, the second edition is not substantially longer 
than the first edition. 

The new material in this second edition has been presented to Electrical 
Engineering classes at the University of Utah with gratifying results. We 
are indebted to these students for their helpful suggestions. We also 
thank John P. Stringham for designing the nomograph given in Fig. 
3.39. The secretarial staff of the Electrical Engineering Department under 
the able direction of Mrs. Joyce Hansen and Mrs. Marian Swenson have 
been unusually helpful and cooperative in the preparation of the manu- 
script. 

We hope you enjoy this new edition. 

Charles L. Alley 
Kenneth W. Atwood 



Prefc 



ace 

TO THE FIRST EDITION 



The recent development of semiconductor devices necessitated the 
addition of one or more courses to an already crowded curriculum in 
Electrical Engineering. In teaching the tube and transistor courses, we 
observed that there was much duplication of both basic theory and 
applications. At about this same time, the s-domain or complex frequency 
concept was introduced into the basic circuits courses. The students 
expressed disappointment that this powerful tool was not utilized in the 
electronic circuits courses also. Therefore, we concluded not only that 
a unified treatment of tube and semiconductor theory was needed, but 
also that the complex frequency concept should be used for electronic 
circuit analysis. The complete response of a system could then be obtained 
rather than the restricted steady-state response. A unified treatment of 
system response to a variety of input signals (such as pulses, step functions, 
sinusoids, etc.) could be achieved, as well as a unification of the tube and 
semiconductor devices. With these objectives in mind, we wrote notes 
which were used in teaching electronics at the University of Utah. These 
notes were used and continually revised for four years before the first 
edition of Electronic Engineering was published. 

As a means of achieving the foregoing objectives, sufficient semi- 
conductor theory is given to provide an adequate understanding of both 
semiconductor and electron tube devices. In general, the basic ideas are 
approached from the point of view of semiconductors. Then, with very 
slight modification, concepts are extended to include the electron-tube 
devices. This broadened application provides better insight into the 
behavior of both semiconductors and electron tubes than is normally 
acquired when each is treated separately. Electron-tube applications are 



x Preface to the First Edition 

considered first, only when this approach provides easier understanding. 

The use of the complex frequency domain leads logically to the use of 
the root-locus technique for analyzing feedback circuits. This approach to 
feedback problems, which provides improved design criteria and visual- 
ization, is used in the discussion of oscillators and negative feedback 
amplifiers. 

The material is organized to provide a consistent development of ideas 
from the basic physical principles of the device to the more complex circuit 
applications. The organization was chosen to provide maximum ease of 
understanding rather than sophistication of presentation. 

To gain maximum understanding of the material in this book, the 
student should have a good foundation in alternating current circuits, 
calculus, and the Laplace transformation. However, since the s-domain 
concept is not used in the first six chapters, the study of circuits and 
mathematics could parallel the study of this book. By accepting some of 
the derived formulas on faith and by using a few transformations in a 
cookbook manner, the essential goals can be achieved by students who 
have had no transformation theory and whose mathematical background 
is otherwise less than ideal. 

Chapter 1 provides a background in electric and magnetic fields. 
This chapter may be omitted if the reader has an adequate knowledge of 
these topics. 

Chapter 2 furnishes the proper background in semiconductor theory. 
If this material has been studied elsewhere, Chapter 2 may also be omitted. 

Chapters 3 to 6 present the principles of operation and methods of 
analysis of typical semiconductor and electron-tube devices. Both 
graphical analysis and equivalent circuits are stressed. 

Chapters 7 to 9 discuss cascaded small-signal amplifiers. In these 
chapters, the s-domain equivalent circuits are used to predict the response 
of typical circuit configurations to various types of excitation. 

Chapters 10 to 13 treat large-signal amplifiers, negative feedback, 
and oscillators. 

Chapters 14 and 15 present the basic principles of modulation and 
detection. Typical circuits are also included. These chapters could be 
omitted without loss of continuity. 

Chapter 16 deals with electronic devices used as switches. Topics such 
as saturation resistance and storage time, which are important in the 
switching mode, are discussed in this chapter. Typical applications such 
as multivibrators are also included. 

Chapter 17 discusses electronic-power supplies, both regulated and 
unregulated. Either Chapters 16 or 17 could be studied in any sequence 
following Chapter 10. 



Preface to the First Edition xi 

An instructor's manual will be provided with problem solutions and 
answers as well as a set of suggested laboratory experiments. A transistor 
manual and a tube manual should supplement the text. In addition, a 
study of the root-locus method from a suitable text such as Introduction 
to Feedback Systems by L. Dale Harris (John Wiley and Sons) would be 
desirable. However, acceptable root-locus plots can be made by following 
the rules in Appendix II. 

We are indebted to the students and staff of the Electrical Engineering 
Department of the University of Utah for the many helpful suggestions 
offered during the revision period. We are especially appreciative of the 
suggestions and comments given by L. Dale Harris and James E. Dalley 
of the Electrical Engineering Department staff. We are also indebted to 
Mrs. Joyce Hansen for the helpful suggestions and supervision of the 
preparation of the manuscript. 

Charles L. Alley 
Kenneth W. Atwood 



r 



Contents 



List of Symbols xix-xxiv 

1 Electron Ballistics 1 

1.1 Properties of an Electron 1 

1 .2 Behavior of an Electron in an Electric Field 2 

1.3 Behavior of an Electron in a Magnetic Field 15 

1.4 Behavior of an Electron in a Combination Electric and Magnetic 
Field 20 

2 Semiconductors 22 

2.1 Atomic Models 23 

2.2 Conduction in a Semiconductor 30 

2.3 Electron Emission 32 

2.4 Charge Carriers in a Semiconductor 33 

2.5 Recombination 39 

2.6 Drift and Conductivity 41 

2.7 Doped Semiconductors 43 

2.8 The Hall Effect 48 

3 Diodes 52 

3.1 The p-n Junction 52 

3.2 The Diode Equation 60 

3.3 Semiconductor Diode Characteristic Curves 63 

3.4 Temperature and Semiconductor Material Effect on the Character- 
istic Curve 64 

3.5 Junction Capacitance 66 

3.6 Junction Breakdown 69 

3.7 The Diode Symbol 71 

3.8 The Tunnel Diode 71 

3.9 High- Vacuum Diodes 77 

xiii 



xiv Contents 

3.10 Vacuum Diode Characteristic Curves 78 

3.11 Symbols for Vacuum Diodes 84 

3.12 Gas Diodes 84 

3.13 Graphical Solution of Diode Circuits 88 

3.14 Diode Equivalent Circuits 91 

3.15 Use of Diode Equivalent Circuits 94 

4 Basic Amplifiers 109 

4.1 Junction Transistors 110 

4.2 The Common-Base Amplifier 1 12 

4.3 Graphical Analysis 114 

4.4 Factors Which Affect a 120 

4.5 Equivalent Circuits 127 

4.6 Input and Output Impedance of the Common-Base Configura- 
tion 140 

4.7 The Triode Tube 143 

4.8 The Triode Equivalent Circuit 147 

4.9 The Grounded Grid Amplifier 150 

J The Common- Emitter Amplifier 154 

5.1 Current Gain of the Common-Emitter Amplifier 155 

5.2 Characteristic Curves 156 

5.3 Transistor Ratings 159 . 

5.4 The Grounded Cathode Amplifier 161 

5.5 Graphical Analysis 161 

5.6 Equivalent Circuits for the Common-Emitter Transistor 170 

5.7 The Equivalent Circuit of a Common Cathode Amplifier 178 

5.8 Biasing Circuits for the Transistor 183 

5.9 Cathode or Self Bias for Tubes 197 

6 Multielectrode Tubes and Transistors 20 J 

6.1 Tetrode Tubes 201 

6.2 Pentode Tubes 205 

6.3 Equivalent Circuits for Pentodes 207 

6.4 Graphical Solution of Pentode Circuits 210 

6.5 Operation of a Pentode from a Single Voltage Source 213 

6.6 Beam Power Tubes 216 

6.7 Tubes with More Than Three Grids 219 

6.8 Field- Effect Transistors 219 

6.9 The Unijunction Transistor 227 

6.10 The Silicon Controlled Rectifier 232 

6.1 1 The Gas Triode or Thyratron 237 



Contents xv 

7 Small-Signal Amplifiers 240 

7.1 The R-C Coupled Amplifier 242 

7.2 Sag and Low-Frequency Response 246 

7.3 The Mid-Frequency Range 251 

7.4 Shunt Capacitance in Amplifiers 254 

7.5 Amplification at High Frequencies 258 

7.6 Diffusion Capacitance 263 

7.7 The Hybrid-Equivalent Circuit for a Transistor 267 

7.8 Gain-Bandwidth Product 272 

7.9 Transformer Coupling 273 

7.10 Frequency Response of a Transformer Coupled Amplifier 277 

7.11 The Common Collector Configuration 285 

7.12 The Cathode Follower 290 

8 Small-Signal Tuned Amplifiers 299 

8.1 Single Tuned, Capacitively Coupled Amplifiers 300 

8.2 Tapped Tuned Circuits 309 

8.3 Inductively Coupled Single-Tuned Circuits 315 

8.4 Double Tuned Circuits 319 

8.5 Neutralization 329 

8.6 The Relationship Between Gain and Stability in a Nonneutralized 
Tuned Amplifier 332 

9 Cascaded Amplifiers 337 

9.1 Gain and Bandwidth Considerations in Cascaded Amplifiers 337 

9.2 db Gain 343 

9.3 Straight Line Approximations of Gain and Phase Character- 
istics 345 

9.4 Cascading Interacting Stages 356 

9.5 D-C Amplifiers 362 

9.6 Noise in Amplifiers 372 

10 Large-Signal Amplifiers 383 

10.1 Classification of Large-Signal Amplifiers 383 

10.2 Distortion and Power Output Considerations 386 

10.3 Graphical Solution of Transistor Circuits for Large Signals 394 

10.4 Efficiency of Class A Amplifiers 399 

10.5 Push-Pull Amplifiers 402 

10.6 Class Box AB Operation 410 

10.7 Thermal Conduction and Thermal Runaway 418 

10.8 Phase Inverter Circuits 422 

10.9 Grid Drive Requirements for Class B 2 (or AB 2 ) Vacuum Tube 
Operation 427 

10.10 Complementary Symmetry Amplifiers 429 



xvi Contents 

11 Amplifiers with Negative Feedback 437 

11.1 The Effects of Feedback on Gain, Distortion, and Noise 438 

1 1.2 The Effect of Negative Feedback on Bandwidth 441 

1 1.3 The Effect of Negative Feedback on Output Impedance 444 

1 1.4 The Effect of Feedback on Input Impedance 447 

11.5 Typical Feedback Circuits 449 

11.6 Stability of Feedback Circuits 454 

1 1 .7 The Characteristics and Stability of Transformer Coupled Feedback 
Amplifiers 460 

11.8 Compensating Networks 463 

11.9 Feedback Design From Gain- Phase (Bode) Plots 467 

12 Large-Signal Tuned Amplifiers 471 

12.1 Basic Behavior of Class C Tuned Circuits 474 

12.2 Graphical Class C Solution for Vacuum Tubes 476 

12.3 Determination of the Tuned-Circuit Parameters 484 

12.4 Class C Transistor Amplifiers 488 

12.5 Self-Bias Circuits for Class C Amplifiers 491 

12.6 The Describing Function 494 

12.7 Efficiency of Class C Amplifiers 498 

13 Oscillator Circuits 503 

13.1 Oscillators with Two-Terminal Active Elements 503 

13.2 Oscillators with Four-Terminal Active Elements 509 

13.3 Typical R-C Oscillators 510 

13.4 L-C Oscillators 516 

13.5 Crystal-Controlled Oscillators 522 

14 Amplitude Modulation and Detection 527 

14.1 Amplitude Modulation 528 

14.2 Modulating Circuits 532 

14.3 The Modulation Process 538 

14.4 Single Sideband Transmission 541 

14.5 Detection of Amplitude Modulated Waves 544 

14.6 Linear Detectors 548 

14.7 Frequency Converters 555 

75 Frequency Modulation 564 

15.1 Modulating Circuits 565 

15.2 Sidebands of the Frequency Modulated Wave 572 

15.3 Interference to FM Transmission 575 

15.4 FM Demodulators 577 



Contents xvii 

15.5 Phase Modulation 583 

15.6 Automatic Frequency Control (AFC) 589 

16 Switching Circuits 593 

16.1 Equivalent Circuits for Electronic Switches 593 

16.2 Capacitance and Charge Storage in Diode Switching Circuits 603 

16.3 Response Time of Electronic Switching Circuits 608 

16.4 Transistor High-Speed Switching Circuits 616 

16.5 The Bistable and Monostable Multivibrator 628 

16.6 The Astable Multivibrator 636 

17 Power Supplies 645 

17.1 Capacitor Input Filters 646 

17.2 Active Filters 651 

17.3 Choke Input Filters 652 

17.4 Polyphase Rectifiers 661 

17.5 Adjustable Output Voltage Rectifiers 666 

17.6 Regulated Power Supplies 671 

Appendix I 685 

Appendix II 688 

Appendix III 697 

Appendix IV 722 

Index 725 



List of Symbols 



A Constant (may have subscripts) 

A Angstrom = 10~ 10 m 

A Area in square meters 

a The turns ratio of a transformer 

a A constant (may have subscripts) 

a Acceleration in m/sec 2 

a Forward current gain parameter of a common base transistor (also h n ) 

<x The low frequency current gain factor 

a' A current gain parameter related to a by Eq. 4.42 

a* The low collector voltage current gain factor 

B Constant 

B Magnetic field intensity in webers/m 2 

B The bandwidth of a single-stage amplifier 

B„ The bandwidth of an n-stage amplifier 

b Ratio of actual coupling to critical or transitional coupling 

b A constant (may have subscripts) 

(3 Current amplification factor (same as h fe ) 

[S Feedback factor 

C Capacitance in farads (may have subscripts) 

C A constant (may have subscripts) 

C Centigrade 

c Velocity of light (2.998 ... x 10 s m/sec) 

y Carrier transport factor 

D Diffusion constant 

D, Distortion with feedback 

xix 



xx List of Symbols 

D n Diffusion constant of electrons 

D„ Open loop distortion (no feedback) 

Z>„ Diffusion constant of holes 

d Length or distance 

d Differential operator 

db Decibels 

A A very small finite amount 

<5 Internal collector efficiency 

3 t The interacting parameter 

E Energy in ergs 

e Base of the natural logarithm 

S Electric field intensity in v/m 

e Permittivity (e r <r„) 

€ r Dielectric constant 

«, Dielectric constant of free space (8.855 . . . x 10~ 1!! f/m) 

F The spot noise figure 

F Force in newtoris 

F a The figure of merit for an amplifier 

/ Frequency in cycles/sec or Hertz 

/ Abbreviation for farads 

f(W) Fermi-Dirac distribution function 

f T The frequency at which the common emitter current gain is one 

£ The damping ratio of a pair of complex poles 

G Giga or 10" 

G A The total gain of a multistage amplifier 

G ( Current gain 

G„ Voltage gain 

G„ Power gain 

GB The gain-bandwidth product 

g Weight in grams 

g Conductance in mhos = 1/r 

g m Transconductance of a vacuum tube 

g mi Transconductance of the screen grid in a vacuum tube 

gtmp Transconductance which indicates the effect the screen grid potential has on 
the plate current 

go The quiescent value of g m 

g e The conversion transconductance of a device 

H Scale factor 

Hz Hertz or cycles per second 



List of Symbols xxi 

h Planck's Constant (6.625 x 10" M j sec) 

h The abbreviation for henries 

hi Input impedance of a device (a second subscript indicates common element in 

configuration) 

h r Reverse voltage amplification factor of a device (may have second subscripts) 

hf Forward current gain of a device (may have second subscript) 

h„ Output admittance of a device (may have second subscript) 

4> T Thermal resistance (may have additional subscripts) 

/ • Current 

Ij Injection current 

I s Saturation current 

I The reverse current of a diode at 300°K 

l c Current in the collector circuit due to thermally generated carriers 

/» The rms noise current 

To differentiate the various currents and voltages associated with tubes and transistors, 
the following notation is used : 

I B A capital letter with a capital subscript is used to denote the quiescent element 
current. 

It, A capital letter with a lower-case subscript is used to denote the rms value of 
the sinusoidal component of current through a given element. This same 
symbol is also used to denote the j-domain current through a given element. 

i B A lower-case letter with a capital subscript is used to denote the total instan- 
taneous current through a given element. 

it A lower-case letter with a lower-case subscript is used to denote the instan- 

taneous time-varying component of current through a given element. 

The subscript letter indicates the particular element which is involved. Thus, B (as 
used in the foregoing examples) denotes the base of a transistor ; C denotes the transistor 
collector; E denotes the transistor emitter; G denotes the grid of a tube. (A second 
subscript consisting of a number may be used to differentiate between grids in a multi- 
grid tube); K denotes the tube cathode; P denotes the tube plate; / denotes the input 
current to a circuit and O denotes the circuit output current. 

/ Current density in amperes/meter 2 

J„ Current density due to electron motion 
J p Current density due to hole motion 
J j Injection current density 
J s Saturation current density 
(-1)* 

K Reference gain of an amplifier 

K Kilo or 10 s 

K A Reference gain of a multistage amplifier 



xxii List of Symbols 

k Coefficient of coupling 

K Constant (may have subscripts) 

°K Degrees kelvin 

k Boltzmann's Constant 1.38 x 10-"j/°K 

k The coefficient of coupling of a coil 

k e The critical coupling of a coil 

k, Kilograms 

L Inductance in henries 

L Length or distance 

L„ The length of the depletion region in n-material 

L v The length of the depletion region in ^-material 

/ Length or distance 

/ Mean-free path in the X direction of an electron 

/„ Mean-free path in the X direction of a hole 

In The natural logarithm of a number 

log The logarithm to the base 10 of a number 

M Mutual inductance 

M Meg or 10 6 

M The modulation index (may have subscripts to indicate the type of modulation) 

m Mass in kilograms 

m Milli or lO" 3 

m The mass of a body at rest 

m e * The effective mass of an electron 

m„* The effective mass of a hole 

H Micro or 10~ 6 

H Charge mobility 

H Voltage amplification factor of a tube 

N Number 

N a Acceptor atom density 

N c The density of states in the conduction band 

N d Donor atom density 

N L Number of electrons which cross a reference plane to the left in time i 

N R Number of electrons which cross a reference plane to the right in time i 

N, Number of states/unit volume 

N x Total number of electrons which cross the relerence surface in time I 

n An integer number 

n Nano or 10~ 9 

n Number of electrons/unit volume 

n The turns of a coil 

n n Number of electrons/unit volume in n-type material 



List of Symbols xxiii 

n v Number of electrons/unit volume in /Mype material 

«, The total number of electrons which cross the reference surface/second 

r) Efficiency (may have subscripts) 

rj The voltage dividing parameter of a unijunction transistor 

co Radian frequency = 27r/(may have subscripts) 

P Power in watts 

P d The power dissipated in a device 

Pi The input power 

P„ The output power 

p Instantaneous power in watts 

p Number of holes/unit volume 

p„ Number of holes/unit volume in n-type material 

p A reference density of holes 

p„ Number of holes/unit volume in /Mype material 

17 A mathematical constant (3.1416 . . .) 

Q The figure of merit for a coil or a tuned circuit 

Q The charge on a body in coulombs 

Q The quiescent operating point of a device 

G The unloaded Q of a circuit 

q The charge on an electron (16.019 ... x 10-*° coulombs) 

R Resistance in ohms 

R,* The total shunt resistance in a circuit 

Re The abbreviation for "real part of" 

R e , The collector saturation resistance 

r Radius in meters 

r, Diode forward resistance 

r b Diode reverse resistance 

r„ Dynamic plate resistance 

r P Static plate resistance 

r, 2 Dynamic resistance of screen grid in a vacuum tube 

p Charge density in coulombs/m* 

S Switch (may have subscript) 

Sj The collector current stability factor 

s Distance in meters. Complex frequency {a +j<o) 

a Conductivity 

T Temperature in °K 

T The time for one period of a periodic function 

T„ A reference temperature usually room temperature (300°K) 



xxiv List of Symbols 

T A The ambient temperature 

T, Junction temperature 

T r The rise time of an amplifier 

Time in seconds 

The rise time of a transistor 

The fall time of a transistor 

The storage time of a transistor 

The delay time of a transistor 

The time for one mean-free path 
t Time constant (may have subscript) 

t The carrier lifetime in a semiconductor 

r x The excess base charge lifetime 

V The potential difference in volts 

V h0 The potential difference across a p-n junction with no external battery 

V h The potential difference across a. p-n junction with an external battery 

V BB Transistor base bias supply voltage 

V cc Transistor collector bias supply voltage 

V<xi Vacuum tube control grid bias supply voltage 

v kk Vacuum tube cathode supply voltage 

V PP Vacuum tube plate supply voltage 

v The velocity in meters/sec 

v„ The rms noise voltage 

v n The mean thermal velocity of electrons 

v p The mean thermal velocity of holes 

Tube and transistor voltages use the same convention as noted for currents. 

W Energy or work in joules 

W a The energy level of the acceptor holes 

W cn Energy at the lowest value in the conduction band in n-type material 

W cv Energy at the lowest value of the conduction band in/7-type material 

W c The lowest energy level in the conduction band 

Wi The energy level of the donor electrons 

Wf Energy of the Fermi-level 

W v The highest energy level in the valence band 

W, The energy width of the forbidden band 

W K The work function of the cathode surface 

w Base width of a transistor 

x Distance along one axis 

Y Admittance in mhos 
Z Impedance in ohms 



Electronic 
Engineering 



1 



Electron Ballistics 



1.1 PROPERTIES OF AN ELECTRON 

Certain properties of an electron are well known. The mass of an 
electron, which is 9.1066 x 10~ 31 kg, is the lightest known finite 1 particle 
of matter. The charge on an electron is 1.6019 x 10~ 19 coulombs. This 
charge is the smallest known unit of electrical charge. In addition, certain 
effects of an electron can be seen. For example, the path of an electron 
under certain conditions can be seen with the aid of a cloud chamber; 
permanent records of some electron paths can be obtained by using 
photographic emulsions or plates. Moreover, certain materials emit 
scintillations of visible light when struck by an electron. Hence it is 
possible to see the actual point of impact of electrons. Since electrons obey 
certain basic laws, their behavior can be predicted and controlled quite 
accurately. In fact, the entire science of electricity and electronics is based 
on man's ability to predict and control the movements of electrons. 

Some properties of an electron are not known. For example, the exact 
size, shape, and color are all unknown. In fact, since the size of an electron 

1 Some "particles" (i.e., photons and neutrinos) are assumed to have zero rest mass. 
No measurable mass has been found to date for these particles. 

1 



2 Electronic Engineering 

is so much smaller than the wavelengths of visible light, it may not be 
proper to associate color with a particle of this size. 

Other properties of an electron are somewhat paradoxical. As an 
example, an electron acts as if it were a small particle in some experiments ; 
whereas in other experiments, electrons act as if they were waves. The 
nature of the experiment determines which property will be the more 
pronounced. However, as far as the electrical engineer is concerned, the 
electron usually behaves as if it were a small particle of matter. 

1.2 BEHAVIOR OF AN ELECTRON IN AN ELECTRIC FIELD 

A basic law of physics states that the force on an object is equal to the 
time rate of change in momentum of the object. If the electron is assumed 
to be a small charged particle, this law can be written 2 as 

F = fe) (1 .D 

dt 
where F = the force, newtons 

m = the mass, kg 
v = the velocity, m/sec 
Now, if the mass is assumed to be constant, 3 Eq. 1.1 can be reduced to 
the familiar form 

F = m — = ma (1.2) 

dt 

where a is the acceleration in m/sec 2 . The equation for the electrostatic 
force on a charged body in an electric field is 

F = Q£ (1.3) 

where Q is the charge on the body in coulombs and & is the electric field 
intensity at the position of the charge in volts per meter. A combination 
of Eqs. 1.2 and 1.3 gives 

= ma (1.4) 



Since acceleration is the time rate of change of velocity, Eq. 1.4 can be 
written as 

m dt 

2 A symbol set in boldface type indicates a vector quantity. 

* The more general case where mass is not a constant will be treated later in this 
section. 



Electron Ballistics 

or 



J m 



^ dt (1.6) 



where v is the velocity in m/sec. Equation 1.6 will apply in all cases 
when mass is constant. The importance of this generality can be realized 
if we note that in some cases the electric field and perhaps even the charge 
are each a function of time. However, a simpler equation can be found if 
the charged body is in a uniform electric field that does not change appreci- 
ably while the electron is in this field. This condition does not restrict the 
use of this simpler equation to fields which do not change with time. 
Because of its small mass and relatively large charge, the electron travels 
at very high velocities. Consequently, as long as the $ field does not 
change appreciably during the time an electron is in the field, the approxi- 
mate formula will be valid. Under these conditions, the $ term of the 
equation may be treated as a constant. If the charge on the charged body 
is constant, and if the mass is assumed to be constant, Eq. 1.6 becomes 



m J 



dt (1.7) 

m 



or 

v = ^t + v„ (1.8) 

m 

where v is the constant of integration and in this case is the initial velocity 
of the charged body at time t = 0. 

Velocity is the time rate of change of distance, so Eq. 1 .8 can be written 
as 

?«^ + *. (!■*> 

dt m 

or 

s = f 2£! dt + fv dt (1.10) 

where s is the distance in meters. Again, assuming Q, S, and m are con- 
stant, Eq. 1.10 becomes 



s = ^- t 2 + \ t + s (1.11) 

2m 

where s is the constant of integration or initial displacement at time t = 0. 

Equations 1.4, 1.8, and 1.11 are very similar to the three corresponding 

equations for a body of mass m in a gravitational field. QS assumes the 



4 Electronic Engineering 

role of the gravitational force. 4 As a consequence, a charged body in an 
electric field "falls" like a body in a gravitational field. 

When a charged body falls through an electric field, the charged body 
loses potential energy as it gains kinetic energy. From the law of con- 
servation of energy, the kinetie energy gained is equal to the potential 
energy lost. Then 

QV=\mv* (1.12) 

or 



- MT (1.13) 

\ m I 



where Q = the charge on the charged body, coulombs 

v — the velocity, m/sec, after falling through a potential difference 

V 
V = the potential difference through which the charged body has 

fallen, volts 
m = the mass of the body, kg. 
This equation will, of course, hold for any charged particle. If the 
numerical values for Q and m are substituted into Eq. 1.13, the equation 
for an electron becomes 

v = 5.93 X 10 5 (K)'^ m/sec (1.14) 

It is interesting to note in passing that if an electron has fallen through 
only 3 x 10~ 7 v, the electron will be traveling at approximately the speed 
of sound in air at sea level. The very large velocities encountered in work 
with electrons can thus be visualized. 

PROB. 1.1. What will be the speed in miles per hour of an electron which has 
been accelerated through 10 v ? How does this compare with 18,000 mph which 
is required to place a satellite into orbit? 

The foregoing equations are based on the assumption that the mass of 
the charged particles remained constant. However, when large velocities 
are encountered, the mass of an object changes. The manner in which 
the mass changes is now considered. 

From Eq. 1.1, 

f = fe) (L1) 

dt 

If mass is assumed to be constant, Eq. 1.2 results. However, if mass is 

4 In the foregoing derivation, the effect of gravity was ignored. Since the electro- 
static force on an electron is usually very much greater than the gravitational force, very 
little error is introduced by ignoring the gravitational force. If large bodic- with small 
charges are considered, both gravitational and electrostatic forces must be considered. 



Electron Ballistics 5 

not assumed to be constant, Eq. 1 . 1 must be written as 

v dv dm .. . 

F = m \- v — (1.15) 

dt dt 

Also, force multiplied by distance in this derivation is equal to kinetic 
energy W. Consequently, Eq. 1.15 can be modified to 

dW=Fds = m — ds + v — ds (1.16) 

dt dt 

Furthermore, since ds/dt is equal to v, Eq. 1.16 can be written as 

dW = mv dv + v 2 dm (1.17) 

Einstein has shown that mass and energy are related by the equation 

E=c 2 m (1.18) 

where E is the total energy, including mass energy, and c is the velocity 
of light (2.998 x 10 8 m/sec). The derivative of Eq. 1.18 may be taken 
to yield 

dE=c 2 dm (1.19) 

This change of energy (dE) is equal to the dW in Eq. 1.17. If this value of 
energy (dE) is substituted into Eq. 1.17, 

c 2 dm = mv dv + v 2 dm 
or 

(c 2 - v 2 ) dm = mv dv (1.20) 

This equation can be rearranged to yield 

dm v . ,. ... 

— = i , dv (1.21) 

m c l — v 

If the mass of the object is m when the velocity is zero, the mass m at 
velocity v can be found by integrating Eq. 1.21 as shown below. 

C m 1 , If" -2v , 

-dm=--\ 2 dv (1.22) 

Jma m 2 JoC — V 

Carrying out the integration, 

In m - In m = -J In (c 2 - v 2 ) + \ In c 2 (1.23) 

which can be written as 



6 

or 



m = 



m 



Electronic Engineering 



(1.25) 



[1 - v 2 /c 2 ] iA 

The relationship between mass and velocity has now been achieved. 
Other useful equations may be established by inserting the value of m 
from Eq. 1.25 into Eq. 1.17. Then, 

dW=Fds= "f , u dv + v 2 d[ m ° . J (1.26) 

However, Fis given in Eq. 1.3 as QS hence, Eq. 1.26 can be written as 

When the integral is taken of both sides of the equation, 

fl[V ds - «*J^f^ dv + -ocj;^^ dv (1.28) 

Since $ ds is the voltage V through which the charged particle has fallen, 
Eq. 1.28 becomes 

QV= mJ-(c 2 - »■)*]"+ m c 



2(c 2 - oV + 



(c 2 



which reduces to 



or 



QV=m c i 



.(-3 M 



- 1 



t> 2 1" 

(1.30) 



QV= (m-m )c 2 
Equation 1.30 can be solved for v. After some manipulation, 

1 



v = c 



1 - 



\ m n c 2 / 



l A 



(1.31) 



When the numerical values for c, Q, and m are inserted in Eq. 1.31, the 
velocity of an electron is 



r i i w 

v = c 1 — _ — - m/sec 

L (1 + 1.966 x 10- S K) 2 J ' 



(1 + 1.966 x 10- 8 K) 2 J 
This value of v can be inserted into Eq. 1.25 to yield 
m = m {\ + 1.966 X \0~ 6 V) 



(1.32) 



(1.33) 



Electron Ballistics 



A plot of velocity vs electron energy 5 is shown in Fig. 1.1. Figure 1.1 
may be divided into three regions. Region 1 extends from to about 
40,000 ev. In this region, m is approximately equal to m and Eq. 1.14 
yields the velocity. In region 2, both mass and velocity change simul- 
taneously. This region 2 extends from about 40,000 electron volts to 



10" 



8l0 8 

E 



10 7 



10 e 







:| : 








--fflti-T^-4-- 






{I 








■ * niii j. 






|| 






HI] .^ 








I 










Region 3 






S — ' 






Region 2 










tt 






T 








Re 


gion 1 






j 
















l 
















t 
















I 
















J 
































1 







1 



10 



10 z 



10 3 



10 4 



10 5 



10 6 



10 7 



Electron energy in electron volts 
Fig. 1.1. Velocity vs energy of an electron as found from Eq. 1.32. 



approximately 3,000,000 electron volts. Equations 1.25 or 1.33 must be 
used to determine the mass, and Eq. 1.32 can be used to find the velocity 
in region 2. Region 3 extends from 3,000,000 electron volts upward. In 
region 3, the velocity is almost equal to the velocity of light and the mass 
can be calculated from Eq. 1.33. Of course, Eqs. 1.32 and 1.33 are 
applicable in any region, but accurate values are quite difficult to evaluate 
in region 1. 

PROB. 1.2. What is the velocity and mass of an electron which has been 
accelerated through 10,000 v? Answer: v =5.93 x W mj 'sec m = 1.02m . 
PROB. 1.3. Repeat Prob. 1.2 for 200,000 v. 

PROB. 1.4. What velocity must an electron have in order to exhibit a mass 
equal to the mass of a proton at rest? What potential must this electron have 
fallen through in order to achieve this velocity? 

Mass of a proton = 1832 mass of an electron. Answer: V = 9.32 x 1CP 
volts v = 2.997 x V0 8 m/sec. 

PROB. 1.5. If an electron is converted into energy, how much energy does the 
electron represent? How much energy is represented by a mass of one gram? 

5 This electron energy is shown as electron-volts where an electron volt is the amount 
of energy an electron receives when accelerated through one volt. 



Electronic Engineering 



Horizontal E|ectron th , 
deflection 




Fig. 1.2. A cathode-ray tube. 

An example will help us visualize the use of the equations just derived. 
In this example, a cathode-ray tube will be used. Figure 1.2 shows the 
principal parts of a cathode ray tube; while typical tubes are shown in 
Fig. 1.3. The electron gun (see Fig. 1.4) produces a narrow beam of 
electrons which have been accelerated through a potential difference of 
several thousand volts. As a result, the electrons are traveling at high 
velocity. These electrons pass between the vertical deflection plates. If a 
potential difference exists between these deflection plates, the electrons will 




Fig. 1.3. Typical cathode-ray tubes. 



Electron Ballistics 



Filaments 



Insulated 
mounting posts 



Glass 
stem 




Seal 
ring 



Cathode 

Accelerating 
electrode 



Deflection 
plates 



Fig. 1.4. A typical electron gun. 

be attracted toward the positive plate and repelled by the negative plate. 
Thus, the beam of electrons is deflected from the axis of the tube an 
amount proportional to the voltage difference between the deflection 
plates. Two sets of deflection plates are provided to obtain both vertical 
and horizontal deflection of the beam. After being deflected, the electron 
beam continues on and impinges on a fluorescent screen, which emits 
visible light when struck by an electron. The cathode-ray tube thus allows 
the visual examination of a voltage waveform. 

Example 1.1. A cathode-ray tube has the dimensions as shown in Fig. 1.5. The 
electrons (in the electron beam) have been accelerated through a voltage V x 
when they enter the space between the deflection plates. The velocity in the 
x direction (along the axis of the tube) is, therefore, 




Fig. 1.5. Configuration for Example 1.1. 



10 Electronic Engineering 

where q is the charge on an electron (16.019 x 10 -10 coulombs). 6 The distance 
traveled by an electron is equal to the velocity multiplied by the time. Hence, 
the time a given electron will be between the deflection plates is given by 



_ L - I m T 

~7 X ~ '\2qTj 



The component of velocity at right angles to the axis of the tube v v can be found 
by using the relationship 

v v = a v t 



Vy "J/* 



The acceleration in the y direction a v can be found from the relationship 

F y = ma y 
But the force in the y direction is 

The electric field intensity can be found from 

where d is the distance between the deflection plates in meters and V y is the 
potential difference between the vertical deflection plates. This relationship 
assumes the electric field between the deflecting plates to be constant. This is a 
good approximation if the dimensions of the plates are large in comparison with 
the distance between them. When the foregoing substitutions are made for a y 
and t, the formula for v y becomes 

_ qV y ll m \W 
"•" md\laV x j 

where v y is the electron velocity in the y direction after passing through the 
deflection plates. This equation can be simplified to the relationship 



d \2mVj 



(1.35) 



Now, if we assume that after leaving the deflection plates, the velocities in the 
x and y directions are constant, we can write 



where /„ is the distance the electron has traveled in the y direction and /„ is the 
corresponding distance in the x direction. In terms of Fig. 1 .5, the foregoing 
equation becomes 

y_ = l 

v y v x 

6 Most cathode-ray tubes have accelerating voltages, V x of 18,000 v or less. Conse- 
quently, the mass can be assumed to be constant. 



Electron Ballistics u 



or 



v, 



y=L-* (1-36) 



»« 



where L and y are defined in Fig. 1.5. When the equalities of v x and v v (Eq. 1.34 
and Eq. 1.35) are substituted into this equation 



d \2mVj 

m 



■ V I 7 

d \2mV, 
y=L- 



or 

LIV, 
y =2dV x 



" (1.37) 



This equation is approximate, since certain simplifications were made 
in the derivation. For example, the fringing effect of the flux field at the 
edges of the plates was ignored. When this fringing is included, the 
equation is about the same as Eq. 1.37, except / is slightly longer than the 
deflection plates. 

The reader may question the validity of Eq. 1.36, which infers that the 
electron has no velocity in the y direction until the electron reaches the 
center of the deflection plates. On reaching the center of the deflection 
plates, however, the electron suddenly acquires the full velocity v v . This 
of course, is certainly not the way the electron would behave. The 
electron actually follows a parabolic path while between the two deflection 
plates. However, if tangents are drawn to the parabolic curve where this 
curve enters and leaves the area between the two deflection plates, these 
tangents will intersect at the center of the deflection plates as shown in 
Fig. 1.5. There is, therefore, no error introduced due to the simplification 
ofEq. 1.36. 

PROB. 1.6. The sensitivity of a cathode-ray tube is defined as the ratio of 
deflection distance (of the electron beam on the screen) to the deflecting voltage 
applied to the deflection plates. What is the sensitivity in cm/v of the cathode- 
ray tube shown in Fig. 1.5 if d = 1 cm, / = 4 cm, and L = 10 in. ? Assume the 
screen is at the same potential as the point where the electron leaves the area 
between the deflection plates. The electron has been accelerated through 2000 v 
when it enters the space between the deflection plates. Answer: 0.0254 cmfvolt. 
PROB. 1.7. How far from the axis of the tube in Prob. 1.6 will an electron be 
when striking the screen if a potential of 10 v is applied between the deflection 
plates? If a particle with the same charge as an electron but with a mass 800 
times as large as an electron is used in this tube, how far from the axis of the 
tube will this particle strike? (Assume 10 v between deflection plates.) Explain. 
What would be the effect of increasing the charge by a factor of 10? 



12 Electronic Engineering 

The derivation of Eq. 1.37 is based on the premise that v x is constant. 
This condition is true only if the potential of the fluorescent screen is the 
same as the potential of the deflection plate field at the point of exit of the 
electrons. If the potential of the screen is different from the potential of 
the field at the point of exit from the deflection plates, the situation becomes 
quite complicated. Fortunately, an approximate relationship can be 
found quite easily. If we assume that the potential varies linearly from the 
center of the deflection plates to the screen, then the average value of v x 
becomes important. In this case 



Vxl + »x2 



(1.38) 



Now v xl is the velocity v x in the original equation (Eq. 1.34) and will 

have the value 

VA 

(1.39) 
m 



V m i 



where V x is again the potential through which the electrons have been 
accelerated when they enter the area between the deflection plates. The 
velocity v xi is the velocity of the electrons when they strike the screen. 
This velocity is given by the equation 

„„,(?ffif (1 .40) 

where V s is the potential difference between the cathode of the electron 
gun and the screen of the tube. Perhaps it is well to note here that the 
velocity of an electron is determined completely by the potential difference 
through which this electron has fallen. Consequently, an electron which 
started from a cathode at zero potential and was accelerated to a velocity 
corresponding to a potential of 100 v will have exactly the same velocity 
as an electron that started at zero potential, was accelerated to a velocity 
corresponding to a potential of 100,000 v, and then decelerated back to 
a velocity corresponding to 100 v above the cathode. The history of the 
electron is therefore not important. All we need to know to determine the 
speed of this electron is the potential at which the electron started (usually 
this is the potential of the cathode) and the potential in which the electron 
finds itself at a given instant. 

When Eqs. 1.39 and 1.40 are substituted back into Eq. 1.38, v av becomes 



\m 1 \ m I 



m ■ (1.41) 



Electron Ballistics 



13 



Now, if this value is substituted back into Eq. 1.36 in place of v x and 
Eq. 1.35 is substituted in place of v y , then 



LIV„ 



y = Liy y (1.42) 

d[V x + (V X V S ) A ] 

This equation is much more useful than Eq. 1.37 because V x is usually not 
equal to V s . Notice that Eq. 1.42 reduces to Eq. 1.37 if V x = V s . Equation 
1 .42 will, of course, be meaningless if V s is equal to zero or is less than zero, 
because the electrons will never strike the screen. 




Fig. 1.6. Action of an electron in passing through a set of deflection plates. 

The electron in the foregoing example appears to have gained some 
energy in being deflected. The source of this energy, however, is not very 
obvious. To explain the energy relations in a cathode-ray tube, refer to 
Fig. 1.6. Assume that no electrons are in the vicinity of the deflection 
plates. A certain charge distribution will then be on the deflection plates 
due to the battery V p . An electron is then injected into the area between 
the plates with a velocity v x which is parallel to the axis of the tube. At 
the point of injection, the electron is in position a of Fig. 1.6. As the 
electron proceeds, it is repelled by the negative plate and attracted to the 
positive plate. Therefore, when the electron leaves the deflection plate 
area, it will be in position b of Fig. 1.6. There has been no force in the x 
direction, so v x is constant. Hovever, the electron now has a velocity in 
the y direction because of the field between the two deflection plates. 
Hence, the electron has more kinetic energy at point b than at point a. 
However, the electron is closer to the positive plate at point b than at point 
a. As a result of the electron being nearer to the positive plate, the charge 
on the two plates is redistributed. In this new condition, more positive 
charge appears on the positive plate, and more negative charge is on the 
negative plate. This movement of charge constitutes a flow of current 
through the battery in such a direction that energy is taken from the 
battery. This energy that is taken from the battery is coupled through the 
electric field to the electron in the space between the two plates. 



14 Electronic Engineering 

As the electron leaves the area between the deflection plates, however, 
another action takes place. The electron at point b of Fig. 1.6 is much 
nearer the positive plate than the negative plate. As a result, the electron 
experiences a decelerating force due to the positive plate and an accelerating 
force from the negative plate. But, since the decelerating force is greater 
than the accelerating force, the electron is slowed down. Also, as the 
electron gets farther from the plates, the charge on the plates will return 
to the charge distribution which existed before the electron entered the 
field, causing the current that flows (due to the change of charge distribu- 
tion) to deliver energy to the battery. Therefore, the electron borrows 
energy from the deflection plates' supply while between these plates. But, 
as the electron leaves the deflection plates, all of the borrowed energy is 
restored to the deflection-plate power source. 

If the screen of the tube is maintained at a potential such that the 
electron is not slowed down after leaving the deflection plates, the electron 
returns its borrowed energy to the deflection plates and obtains new energy 
at the same rate from the field of the screen. The electron, therefore, 
obtains no net energy from the deflection plates. The deflection plates 
do change the direction of travel of the electron, however. 

The previous work assumed that the electric field remains essentially 
constant as a given electron travels through the field. As already noted, 
this condition is met in practical situations except when the rate of change 
of the electric field is very fast. However, if the electric field changes very 
rapidly, the voltage between the deflection plates will be different when the 
electron leaves the deflection plate area than when this electron entered 
the area. The electron can then acquire a net energy from the field of the 
deflection plates. The source supplying energy to the deflection plates 
must be capable of furnishing this energy. This loss of driving energy 
causes difficult problems when frequencies in the microwave band (several 
thousand megacycles) are being used. 

PROB. 1.8. How far from the axis of the tube (Fig. 1.7) will the electron strike 
the face of the tube ? 

Potential of cathode = -2000 v 

Potential of plate 1 = +10 v 

Potential of plate 2 = 10 v 

Potential of final electrode in electron gun = v 

Potential of screen = +2000 v. Answer: y = 0.311 cm. 

PROB. 1.9. Repeat Prob. 1.8 with a screen potential of Ov. Does this condition 
cause the tube to be more or less sensitive? Explain. 

PROB. 1.10. When the voltage difference between two deflection plates 
becomes large enough, the electron beam will strike one of the deflection plates. 
At what deflection plate voltage does this condition occur in the tube of Prob. 
1.8? Answer: V y = 445 volts. 



Electron Ballistics 



15 




Fig. 1.7. Configuration for Probs. 1.8, 1.9, and 1.10. 

PROB. 1.11. How much time is required for an electron to pass between two 
deflection plates 3 cm long? Assume that the electron has been accelerated 
through a potential V s of 2000 v before entering the area between the deflection 
plates. If a signal of 20 sin 2wl0 6 / is applied to the deflection plates, what is 
the maximum voltage change which can occur while the electron is between the 
deflection plates? 

1.3 BEHAVIOR OF AN ELECTRON IN A MAGNETIC FIELD 

An electron in a magnetic field experiences a force only if the electron 
has a component of velocity at right angles to the flux lines of the magnetic 
field. This principle is illustrated in Fig. 1.8. The component of velocity 
v x which is perpendicular to the flux lines B z produces the force F v . A 
velocityinthe +y direction produces a force in the — redirection. However, 
a velocity in the z direction (parallel to the magnetic force) produces no 



Uniform magnetic field B z (out of 

• ••••••/•••• 

■ Path of electron 

F y = force due to A • 
magnetic field! # 

v * ' .. ' . .. « > — ~~ 

Velocity of ••••••••••• 

electron 

Fig. 1.8. Action of a magnetic field on an electron. 




16 Electronic Engineering 

force. The magnitude of the force can be found from the relationship 

F y = B/p>„ (1.43) 

where F y = force in the y direction in newtons. 

B z = magnetic flux density in the 2 direction in webers per square 
meter. 
q = charge on the electron (16.019 X 10 -20 coulombs). 
v x = velocity in the x direction in m/sec. 
Similar relationships can be derived for forces in the x and 2 directions. 
A general expression can be written in vector form 7 as 

F=g(vXB) (1.44) 

where Q is the charge on any charged body and includes the sign of the 
charge. F, v, and B are vector quantities for force, velocity, and magnetic 
flux. 

The force is always at right angles to the velocity (Fig. 1 .8). In addition, 
this force is constant for a constant velocity and constant magnetic flux. 
Under these conditions, the path of an electron will be circular. The force 
due to Eq. 1.44 is tending to pull the electron toward the center of the 
circle. In contrast, the centrifugal force of the electron tends to pull the 
electron away from the center of the circle. This centrifugal force is equal 
to 

F = ^ (1.45) 

r 
where F = the force, newtons 

m = the mass of electron, kg 
v = the velocity of electron, m/sec 
r — the radius of curvature of the electron path, m 
These two forces (Eqs. 1.44 and 1.45) must be in equilibrium and thus can 
be equated as shown by Eq. 1 .46. If the velocity is perpendiculai to the 
magnetic field, 

D mt;2 t, AH\ 

Bqv = (1-46) 

r 
From this equation, 

mu ,. ._. 

r = — (1.47) 

Bq 
or 

v = ^ (1.48) 

m 

7 (v x B) is read as v cross B and does not mean v "times B. This cross product means 
the component of v which is at right angles to B multiplied by B. The resultant force is 
at right angles to both the component of v (which is at right angles to B) and to B. 



Electron Ballistics 17 

In making one complete circle, the electron travels a distance s as given 
below. 

s = 2-rrr (1.49) 

The time required to make a complete circle T is given by 

T= S - (1.50) 

v 

When the values of s and v from Eqs. 1.49 and 1.48 are substituted into 
Eq. 1.50, 

Irrrm 
Bqr 
or 

T=— (1-51) 

Bq 

For an electron, m and q are constants (for velocity « speed of light), so 
the time for one circle is dependent only on the magnetic flux density. 
When the values for the charge and mass of an electron are substituted 
into Eq. 1.51, 

T=^xl0- 12 sec (1.52) 

B 

Magnetic fields are used in many applications. For example, magnetic 
deflection of electron beams in cathode-ray tubes (especially in television 
receivers) is widely used. Also, magnetic fields can be used for focusing 
electron beams, and they are even used in the generation of microwave 
frequencies. Furthermore, charged particles are accelerated to very 
high velocities by "cyclotrons." This latter application will be con- 
sidered as an example of the use of magnetic fields. 

A cyclotron is constructed as shown in Fig. 1.9. Two hollow D-shaped 
conductors are arranged as shown. An electron emitter or gun is located 
inside the hollow D's and very near the center of the cyclotron. A uniform 
magnetic flux is maintained throughout the area of each D. Finally, a high 
voltage a-c source is connected between the two D's. Now, assume a 
group of electrons is emitted from the electron gun just as D no. 1 is 
maximum positive and D no. 2 is maximum negative. These electrons are 
accelerated into the D no. 1 space. Since the electrons now have a velocity, 
they will be deflected into a circular path as shown in Fig. 1.9. If the 
frequency of the a-c supply is correct, the electrons will arrive at the gap 
between the two D's (position b) just when D no. 2 is maximum positive 
and D no. 1 is maximum negative. The electrons will again be accelerated 
between the two D's. Since the electrons will now have a higher velocity 



18 



Electronic Engineering 



than they had from a to b, the radius of the electron path will be greater. 
Similarly, the electrons will arrive at point c just in time to be accelerated 
again. The electrons continue this action for many cycles. Each time the 
electrons cross the gap between the D's, an increase of velocity and radius 
of the path results. When the velocity is sufficiently high, the electrons 
are removed from the magnetic field and travel in a straight line to the 
desired target. Since a group of electrons can be released once each cycle, 



flno. 1 




Uniform magnetic field 
over D areas 

, — f/v)— . Hollow D 

||/" pieCe 



— 7^\ 

/ V Plane of 

in S electron Da' 



Electron gun -^ electron paths 

Side view 



Dno. 2-^X 

Electron 
path 

Top view 

Fig. 1.9. A simplified cyclotron. 

the output of a cyclotron is a periodic pulse of high-velocity particles. 
Whereas the foregoing analysis has used an electron as the charged particle, 
most cyclotrons are used to accelerate protons, hydrogen nuclei, or other 
large particles. However, the analysis is the same. 

Example 1.2. A cyclotron has the configuration shown in Fig. 1 .9. The maximum 
voltage from the a-c source is V m . If mass is assumed to be constant, the elec- 
trons will have a velocity of 



v = 5.93 x l(P(V m y<< m/sec 
after the first passage through the gap. From Eq. 1.46 

r = — = 3.37 x 10- 6 ^— meters 
Bq B 

In the curve from b to c (Fig. 1.9), the velocity will be 

v = 5.93 x 10 5 (2K m )^ 

and the radius Of the path will be 

r = 3.37 x 10- 



(1.14) 



(1.53) 



,(2V m )X 



B 



19 
Electron Ballistics 

In general, the velocity after passing through the gap n times is 

v n = 5.93 x 10 5 («K m )^ (1-54) 

and the radius of the path for this velocity will be 

r= 3.37x10-"^^ 0-55) 

The frequency of the high voltage source can be found by noting 

r = !H x 10-12 sec (1.52) 

B 

and, 

1 B 

f = -= x 10 12 cvcles/sec (1-56) 

J T 35.5 J 

Cyclotrons of the type just described are used only for accelerations up 
to about 0.1 of the velocity of light. When greater accelerations are 
required, the change of mass must be considered. As noted in Eq. 1.51, the 
period becomes longer as the mass increases. Some cyclotrons have been 
built which are frequency modulated (the frequency is changed) to allow 
for the change of mass. At the time a given group of electrons (or other 
particles) leaves the emitter (or particle source) the frequency of the high 
voltage oscillator is high. As the mass of the particles increases, the fre- 
quency of the oscillator decreases. The given group of particles, therefore, 
always arrives at the gap between the Z>'s just as the field is a maximum. 

PROB 1 12. (a) How many times must an electron pass through the gap of a 
cyclotron before achieving a velocity of 0.1 times the velocity of light? Assume 
the high voltage oscillator produces 1000 v peak, (b) How many times must a 
hydrofen ion (composed of one proton) pass through the gap of this cyclotron 
before achieving a velocity 0.1 times the velocity of light? 
PROB. 1.13. (a) If the magnetic field strength of Prob. 1.12 is 1.53 x KT 2 
webers/m 2 , how large must the C's be to accommodate the largest radius path? 
(b) To what frequency should the oscillator be set? Answer: d = 40.8 m, 
f = 2.34 x l(fi cps. 

PROB 1.14. A cyclotron is to be designed to furnish electrons with an energy 
of 1 000 000 ev. A high voltage oscillator with an output of 10,000 v is to be 
used. Use a magnetic flux density of 1.5 x 10" 2 webers/m 2 . (a) Find the size 
of D required, (b) What is the frequency of the oscillator at the start of an 
electron bunch? (c) What is the frequency of the oscillator at the time the 
electrons leave the Z>? (d) How many bunches per second can this cyclotron 
produce? (Assume the frequency changes at a linear rate.) 
PROB. 1.15. A magnetic field exists as shown in the sketch of Fig. 1.10. Two 
electrons enter this field at point A with a velocity of 10 7 m/sec. The velocity of 
electron a has the direction shown where a = \°. The velocity of electron b 
has the direction shown where /? = 1°. (a) How far from A will electron a again 



20 Electronic Engineering 




Uniform 
flux density 



-*~ B=10- 2 webers/m 2 



Fig. 1.10. Configuration for Prob. 1.15. 

cross the axis ? (Hint: Break the velocity of the electron a into two components, 
one component parallel to the magnetic flux lines and one component perpen- 
dicular to the magnetic flux lines.) (b) What path will electron a follow? (c) 
How far from A will electron b again cross the axis ? (d) What use does this 
problem suggest for magnetic fields ? 

1.4 BEHAVIOR OF AN ELECTRON IN A COMBINATION 
ELECTRIC AND MAGNETIC FIELD 

When an electron is in an area that contains both an electric and a 
magnetic field, the force on the electron is the vector sum of the force due 
to the electric field and the force due to the magnetic field. These forces may 
either tend to reinforce each other or tend to cancel each other, depending 
on their respective directions. 

It is possible for one force to completely cancel another force. Such 
a condition is shown in Fig. 1.11. The magnitude of the force due to the 
electric field, F E , can be found from Eq. 1.3. 

F E = QS (1.57) 

The magnitude of the force due to the magnetic field, F M , can be found 
from Eq. 1.43. 

F M = BQv (1.58) 

If these two forces are equal and opposite in direction, the electron will 

+ Charge 



/ 


'* 


X 


X 


X 


X A F E x 


X 


X 


X 


Parallel 


X 
X 


X 
X 


X 


X 


X 


X 


Uniform magnetic field (into page) 

XXX 


plates 
V 


X 


X 


X 


X 


X 


x^ 


\ x 


\ 


^ x 


X 


X 


X 


x' 


%* 


X 


X 


^ Path of electron 
X 



-Charge 
Fig. 1.11. Cancellation of force due to an electric field by a magnetic field. 



Electron Ballistics 



21 



continue in a straight line. The two forces can be equated to give 



or 



Qg = BQv 



B ~ 



(1.59) 



where £ = the electric field intensity in v/m 

B = the magnetic flux density in webers/m 2 

v = the velocity of the electron in m/sec 
If the velocity of a charged particle is less than this v, the particle will be 
pulled toward the positive plate. In this case, F E is greater than F M . If 
the velocity of the charged particle is greater than v as given by Eq. 1.59, 
the particle will be forced toward the negative plate. This type of con- 
figuration is known as a velocity filter. Only particles with a given velocity 
will be passed; all other particles will be captured by the two plates. A 
filter of this type may be used on a cyclotron to remove the particles in the 
outermost orbit from the magnetic field. 

PROB. 1.16. Design a velocity filter to remove electrons from the cyclotron of 

Prob. 1.13. The spacing between plates must be 1 cm. What voltage must be 

applied to these plates ? Answer: V = 4590 volts. 

PROB. 1.17. Repeat Prob. 1.16 for the cyclotron of Prob. 1.14. 

PROB. 1.18. An electron enters the space between two deflection plates with a 

velocity of 13.27 x 10 6 m/sec. If the configuration of the deflection plates is as 

given in Fig. 1.12, how far from the input end of the deflection plate will the 

electron strike the deflection plate? 



+ 10v 



Electron- 



13.27 x 10 6 
m/sec 



^ 



2 cm 



-lOv 



-y = 



20 v 



i = 



Fig. 1.12. The configuration for Prob. 1.18. 

PROB. 1.19. How far will the electron of Fig. 1.12 have traveled in the y 
direction when it has traveled 10 cm in the x direction? 



2 



Semiconductors 



Modern engineering must be based on an understanding of the basic 
physical properties and processes of materials. A thorough understanding 
of engineering materials may be obtained only from a serious study of 
advanced subjects such as atomic physics, physical chemistry, and quantum 
mechanics. The development of new devices is predicated on the intelligent 
application of these principles to the problems of modern technology. 
This frequently requires the cooperative effort of physicists, chemists, and 
engineers. 

The effective use of solid-state electronic devices requires at least a 
superficial knowledge of the physical principles on which they operate. 
Such a knowledge should also allow the engineer to understand and 
effectively use the many devices which will almost certainly be developed 
in the future. 

The purpose of this chapter is to present some of the physical principles 
of semiconductor material. The treatment will, of necessity, be superficial, 
since a thorough treatment is clearly beyond the scope of this book. It is 
hoped, however, that the reader will acquire some mental images and basic 
relationships which are sufficiently correct to permit him to intelligently 
use the available devices. Consequently, physical models will be used at the 

22 



Semiconductors 23 

expense of mathematical rigor in an effort to make the otherwise intangible 
principles somewhat concrete. For example, the Bohr model of the atom 
will be used in favor of the quantum theory because the particle theory of 
matter is easier to visualize than its wave nature. Models and analogies 
will be borrowed frequently from the literature. An attempt will be made 
to give proper credit, but this will not always be possible because the 
original sources may be unknown or the evolutionary modifications of the 
original may make it impractical, if not impossible, to give the deserved 
credit to all the contributors. 

2.1 ATOMIC MODELS 

The Bohr model of the atom places an orderly arrangement of electrons 
in elliptical orbits around a nucleus of protons and neutrons. This concept 
was contributed primarily by Bohr during the early part of this century. 
The theories developed from this model agreed so well with experimental 
evidence that the measuring instruments and techniques of that period were 
unable to detect any difference between the predicted and actual behavior 
of matter. However, as the measuring instruments and techniques 
improved, it became evident that discrepancies existed. 

The discovery and investigation of the photoelectric effect provided 
convincing evidence that light has a dual nature. Some effects, such as 
diffraction, can be explained only on the premise that light behaves as a 
wave. Other phenomena, such as the mechanism of photo emission, can 
be explained only on the basis that light consists of bundles of energy or 
particles known as photons. Later it was shown by Davisson and Germer, 
by the use of diffraction techniques, that electrons in motion also have 
the properties of a wave. This rather startling discovery led DeBroglie, 
Schrodinger, and others to the development of a wave theory for all 
matter. This wave mechanics or quantum mechanics eliminates certain 
discrepancies inherent in the Bohr model. An atomic model which 
accurately represents the wave mechanics concept is difficult to visualize, 
however, because the particles are not so discrete and well defined. For 
example, probability densities which are dependent on boundary conditions 
replace the solid charged satellites of the Bohr model. For this reason, the 
Bohr model will be used, as stated previously, to aid the visualization of 
the basic principles. 

The Bohr model diagrams of a few atoms of interest are shown in Fig. 
2.1. The simplest atom is hydrogen which has one planetary electron and 
a nucleus consisting of one proton. The tetravalent atoms are of con- 
siderable interest to the semiconductor industry. The simplest of these is 
carbon, which has six planetary electrons and six protons in the nucleus. 



24 Electronic Engineering 

The two electrons nearest the nucleus form what is known as a "closed 
shell." That is, only two electrons can be accommodated at that energy 
level. The four outer electrons are known as valence electrons and they 
determine the chemical activity of the material. These valence electrons 
were originally thought to occupy the same energy level. However, closer 
investigation has revealed that the valence shell is split into at least two 
levels, with two electrons in the lower level or subgroup. The outer level 




Hydrogen 



Carbon 



Silicon 
Fig. 2.1. Bohr model diagrams of a few atoms. 

would require four additional electrons to fill the shell. This behavior 
may be observed from the silicon diagram which, like carbon, has four 
valence electrons but has a filled shell between the inner shell and the 
valence shell. Silicon's chemical properties are similar to those of carbon. 
Another element in this group is germanium, which has thirty-two orbital 
electrons including four valence electrons. Germanium has two filled 
shells between the valence electrons and the inner shell. 

The energy of the orbital electrons may be determined from the laws of 
mechanics. The hydrogen atom will be used as an example. If the single 
orbital electron is removed to infinity, the atom is said to be ionized. The 
electron at infinity is considered to be at rest and to have a potential energy 
equal to zero. This is an arbitrary choice of reference for potential energy, 
but it is a convenient and widely accepted one. If the electron is given a 
very slight nudge in the direction of the nucleus, the electron will travel 
toward the nucleus and will be accelerated by the electrostatic force 



F = 



4ner 2 



(2.1) 



Semiconductors " 

where e is the permittivity and r is the distance between the charges. 
The potential energy at any given point is the work required to bring 
the charge from the reference to the point in question. If the point is a 
specific distance r from the nucleus and the reference is at r = oo, then the 
potential energy is 



„,_f -*-,*_-£- (2.2) 

J» 4ire,,r 47re„r 



where e„ is the permittivity of free space. The negative sign indicates 
that the potential energy of the electron decreases as it moves toward 
the nucleus. In contrast, the kinetic energy increases as the electron moves 
toward the nucleus because the electron is being accelerated by the force. 
As noted in Chapter 1, the electron can be considered as "falling" toward 
the nucleus because of the similarity to a mass falling in a gravitational 

field. 

If the electron has any component of velocity or force normal to the 
radial direction r, the electron will go into an elliptical or circular orbit 
around the nucleus. For mathematical simplicity, it will be assumed that 
the orbit will be circular with radius r. In order to have equilibrium, the 
electrostatic force between the electron and nucleus must be equal to the 
centrifugal force. 

F = JL- = ™ 2 (2.3) 

4ir€ v r z r 

where v is the tangential velocity of the electron. 

Then, using Eq. 2.3, the kinetic energy of the electron is 

W k = \mv* = /- (2.4) 

It is interesting to note by comparing Eqs. 2.4 and 2.2 that the kinetic 
energy of the electron in orbit is only half as great as the decrease in 
potential energy from the reference level. The law of conservation of 
energy then requires that half of the potential energy change be transferred 
to some other type of energy, since only half was transferred to kinetic 
energy. Observation of an ionized gas will reveal that electromagnetic 
radiation accounts for this energy transfer. This radiation is frequently in 
the visible spectrum and is the mechanism which produces the light in 
neon tubes or similar devices. 

The total energy level of the electron is 

W=Wj>+Wk =-^+^-=-^- (2.5) 

47T€„r 877€„r 8ir€ v r 



26 Electronic Engineering 

The period of the orbital motion is 

T=^ (2.6) 

v 

A frequency equal to 1/Tis associated with the electron. There can be no 
electromagnetic radiation at this frequency as long as the radius remains 
constant because there can be no change in the total energy of the electron. 
Radiation or absorption of energy can only be associated with a change in 
radius and thus a change in energy level. 

It has long been known that the atom can be stable only when the 
orbital electrons have certain discrete energy levels. The energy differences 
between these permissible levels indicate the amounts of energy or size of 
energy packets that can be either radiated or absorbed by the atom. In 
1900, Max Planck showed that the relationship between the frequency/ 
of the electromagnetic wave, either radiated or absorbed, and the discrete 
energy packet is 

W=hf (2.7) 

where/ = the frequency of the electromagnetic radiation 

h — Planck's constant 6.625 x 10~ 34 j sec 
The lines on a spectrograph of a specific material could then be used to 
determine the size of the energy packets for a given material. When 
Planck made his discovery, the electron had not been discovered so it 
could not be known that the energy packets were the difference of per- 
missible electron energy levels, or W = W x — W 2 . 

The development of wave mechanics provided an explanation for the 
discrete energy levels of the electron. If the electron in motion has the 
characteristics of a wave, the electron in a stable orbit must behave as a 
wave which has an integral number of wavelengths in one circumference of 
its orbital path. Otherwise, the electron waves would tend to cancel one 
another by interference, as illustrated in Fig. 2.2. The peaks and troughs 
of the wave are represented by the solid line. In 1924, DeBroglie postulated 
that the wavelength A of *he matter waves is 

A = — (2.8) 

mv 

where the wavelength A is the distance between the peaks (or troughs) of a 
wave (A = vjf). Davisson, Germer, and Thompson later demonstrated 
the wave nature of electrons by obtaining interference patterns from 
crystal diffraction gratings placed in a stream of moving electrons. If we 
accept the postulate that there must be an integral number of wavelengths 



Semiconductors 



27 





M (b) 

Fig. 2.2. Wave characteristics of an electron in (a) a stable orbit, (6) an unstable orbit. 

in one circumference of the orbital path, it follows that 



, „ nh 

tiA. = 2irr = — • 
mv 



where n is an integer. 

Using Eq. 2.4 to solve for v in terms of r, we see that 



\4ire,,rm/ 



Then 



2-rrr = 



nh(Aire v rmy 
mq 



Solving for the permissible values of r = r„ 



■nmq 



The stable values of energy, from Eq. 2.5 are 

w £f-=^™*l 

877€„r„ 8€„ 2 hV 



(2.9) 

(2.10) 
(2.11) 

(2.12) 

(2.13) 



It should be remembered that the foregoing derivations are for the 
simplest configuration of a single hydrogen atom. The object was merely 
to illustrate the fact that the electrons associated with an atom may have 
only discrete values of energy in the stable state. This fact is illustrated in 
the energy diagram of Fig. 2.3. It should be evident that this basic 
principle applies to all atoms. 

The precise energy levels of the electrons of an individual atom are 
modified by the presence of other atoms in near proximity. For example, 



28 



Electronic Engineering 



an electron associated with one atom of a group is influenced by the charge 
distributions of the neighboring atoms in the group. A particular electron 
in a tightly packed group of hydrogen atoms experiences a force caused by 
a neighboring atom, provided that the electron in question is not equi- 
distant and midway between the electron and nucleus of its neighbor. 
The resulting force would change both the potential and the kinetic 




Fig. 2.3. Electron energy levels of a hydrogen atom. 

energies of the electron. This effect applies to all electrons in the group. 
Thus, when the atoms of a particular element are spread far apart, the 
permissible energy levels of one atom are precisely the same as any other 
atom; but, as the spacing between the atoms is decreased, the energy 
levels of each electron are increasingly influenced by the charge distribu- 
tions of the neighboring atoms. The magnitude of the influence depends 
not only on the spacing but also on the location of the electron within 
the group. Consequently, the discrete electron energy levels of the isolated 
atoms change into energy bands when other atoms are in close proximity. 

It is meaningful to talk about the spacing between individual atoms only 
when they form a crystalline solid. Then the atoms have an orderly 
arrangement with a rather precise spacing. This spacing between repre- 
sentative atoms in the crystal is known as the lattice constant. In general, 
the semiconductor materials used in electronic devices are crystals, and 
one of the manufacturing problems is to obtain a very pure, regular 
crystal. 

Figure 2.4 shows the energy bands as a function of lattice constant for 
tetravalent crystals such as diamond (carbon), silicon, and germanium. 



Semiconductors 29 

The energy bands between the permissible bands are known as forbidden 
bands. The actual valence electrons of one atom are indicated by a small 
circle with a negative sign inside. This condition is called a filled energy 
state. The small empty circles represent energy states which are available 
for but not occupied by electrons in a particular band. For example, there 
are eight energy states available in the valence shell of a carbon atom but 



Conduction 
band 




Ge Sn 
Atomic spacing - 

Fig. 2.4. Energy bands of a tetravalent crystal as a function of lattice spacing. 

only four of them are actually occupied or filled. When the temperature is 
absolute zero, the electrons are in the lowest possible states. This condition 
is assumed in Fig. 2.4. The energy levels below the valence band are not 
shown because they are all filled and do not contribute to the electrical 
characteristics of the material. 

Figure 2.4 shows that as the lattice spacing decreases from its initial 
large value, a single band is first formed. This band is a combination 
valence and conduction band. As the spacing is further reduced, the band 
divides into separate conduction and valence bands with a forbidden band 
between. In this case, at 0°K the valence band is completely filled and the 
conduction band is completely empty. It would be impossible to adjust 
the atomic spacing of a particular crystal arbitrarily. However, the lattice 
constant of crystals of different elements are different, so different elements 
having similar valence shells such as the tetravalent group could be used 
to provide variation of lattice constant in discrete steps. The spacing 



30 Electronic Engineering 

between nuclei would naturally increase with the complexity of the atom 
because of the larger atomic diameter. Typical energy-band configurations 
for carbon, silicon, germanium, tin, and lead are indicated in Fig. 2.4. 



2.2 CONDUCTION IN A SEMICONDUCTOR 

The conduction process depends on the availability of charge carriers 
which are able to move in the presence of an electric field. The acceleration 
of the carrier by an electric field increases the carrier's total energy. This 
increase of energy level can be accomplished only if an unfilled state is 
available at a slightly higher energy level. Referring again to Fig. 2.4, it is 
evident that lead is a good conductor because there are unfilled energy 
states in the same energy band as the valence electrons. On the other hand 
germanium, silicon, and diamond crystals are insulators at 0°K because 
their valence bands are completely filled and their conduction bands are 
completely empty. A comparatively large amount of energy would then be 
required to elevate a valence-band electron through the forbidden band to 
the conduction band. A very strong electric field could elevate an electron 
to the conduction band, but this would be a process known as the "voltage 
breakdown" rather than a normal conduction process. When the tempera- 
ture is raised above absolute zero, the crystal atoms acquire vibrational 
kinetic energy. Some of the valence electrons may acquire sufficient 
energy to jump from the valence band to the conduction band. The 
crystal would then be a conductor of sorts because there would be carriers 
in the conduction band and also unoccupied states in the valence band. 
The conductivity would depend on the number of carriers in the conduction 
band. Thus, the higher the temperature, the better the conductivity. The 
crystal is called a semiconductor because near room temperature it has 
electrical conduction properties between those of a conductor and an 
insulator. 

The conduction due to the electrons in the conduction band is a different 
process than the conduction due to the unfilled states left in the valence 
band. The latter is called conduction by "holes." In the intrinsic or pure 
semiconductor material, there are as many holes as there are free electrons, 
as the former is created by the latter. An analogy (which is attributed to 
Shockley) might be used to illustrate the conduction processes. A parking 
garage with two floors has the lower floor completely filled with auto- 
mobiles and the upper floor completely empty. Under these conditions 
there can be, no movement of automobiles on either floor. If one auto- 
mobile is elevated from the lower to the upper floor, there can be motion of 
automobiles on each floor. The auto on the upper floor may move freely 
over comparatively large distances. In contrast, the motion on the lower 



Semiconductors ■" 

floor is accomplished by moving one car at a time into the available space. 
Hence, an observer near the ceiling of the first floor would see the "hole" 
move rather than the automobiles. The hole would have less mobility 
than the auto on the upper floor. Nevertheless, both would contribute to 
the total motion. 

A qualitative examination of the crystal structure may give additional 
insight into the behavior of the valence-band electrons. Under carefully 




Fig. 2.5. Two-dimensional representation of covalent bonds in a tetravalent crystal. 

controlled conditions, the molten material solidifies into an orderly 
arrangement of atoms. A two-dimensional symbolic diagram of this 
arrangement for a tetravalent material is shown in Fig. 2.5. Covalent 
bonds exist between each pair of atoms and hold the crystal together. In 
this manner, the atoms share their valence electrons, so each atom feels 
that it has realized its ambition to fill its valence shell. In fact, the total 
electron energy is less in the crystalline arrangement than it would be in a 
haphazard arrangement. Figure 2.4 show that the energy of the valence 
electrons is reduced as the crystal is formed. This reduction in energy 
accounts for the "latent heat of fusion" which occurs when a liquid freezes. 
In Fig. 2.5, each circle represents the nucleus and the closed shells of a 
tetravalent atom. Consequently, the net positive charge of this inert group 
is equal to the charge of the four valence electrons. Each minus sign 
represents a valence electron and the double lines represent the covalent 
forces or bonds which hold the atoms together. Thus, the diagram 



32 Electronic Engineering 

represents the condition of the entire intrinsic germanium crystal at 0°K. 
As the temperature is raised above 0°K, the energy of the crystal atoms 
is increased and they vibrate about their equilibrium positions. If an 
electron receives sufficient energy, it may escape from the valence position 
or break the covalent bond and drift through the crystal lattice. The 
electron has then received the energy required to jump from the valence 
band to the conduction band. This energy is about 0.78 electron volt for 
germanium and 1.2 electron volts for silicon at 0°K. The electron may 
receive this energy by heat, light, particle bombardment, or intense electric 
field. 

Conduction in the conduction band results from the movement of the 
"free" electrons under the influence of an electric field; this movement 
is hampered only by collision with the stationary lattice atoms and local 
atomic fields. The mechanism of the hole conduction is merely the move- 
ment of an electron from one covalent bond to a nearby missing covalent 
bond under the influence of an electric field. 

2.3 ELECTRON EMISSION 

The process of electron emission which is utilized in electron tubes 
requires that electrons escape from the surface of a conductor into the 
surrounding gas or vacuum. The energy of the electron must then be 

TABLE 2.1 
Commonly Used Cathode Materials 

Material W w (ev) Melting Point (°K) 



Cesium 




1.81 


299 


Copper 




4.10 


1356 


Nickel 




4.60 


1725 


Thorium 




3.40 


2118 


Tungsten 




4.52 


3643 


Thorium on 


tungsten 


2.60 




Rare-earth oxides 


1.00 (typical) 





increased from the conduction-band level to essentially the zero-reference 
level at infinity, which is known as the vacuum level. The energy difference 
between the highest occupied state at 0°K and the vacuum level is known 
as the work function of the material. The electrons may receive sufficient 
energy for emission from heat, light, bombardment by particles, or intense 
electric fields. These types of emission are known as thermionic emission, 



Semiconductors 



33 



photoemission, secondary emission, and high field emission, respectively. 
The emitted electrons are free charge carriers in the space surrounding the 
emitter. These electrons drift in the presence of an electric field and thus 
cause the space to be a conducting medium. 

Some commonly used cathode materials are listed in Table 2.1. The 
work functions W w and melting points are also given. An ideal thermionic 
emitter would have a low work function and a high melting point. The 
mechanical properties of the material also need to be considered. Com- 
monly used thermionic emitters are the rare-earth oxides, tungsten, and 
thorium on tungsten. Cesium is sometimes used as a photo emitter. 

2.4 CHARGE CARRIERS IN A SEMICONDUCTOR 

The conductivity of a semiconductor depends on the charge-carrier 
density. Therefore, conductivity depends on the temperature, as pre- 
viously discussed. In order to predict the conductivity characteristics of 
a material, a relationship between charge-carrier density and temperature 
must be developed. At temperatures above absolute zero, some valence 
electrons break their covalent bonds and become carriers. At the same 
time these electrons leave holes which act as positive carriers. In addition 
there is a continual recombination of electrons and holes. Therefore, the 
total number of carriers will reach a steady-state value at a particular 
temperature. 

It would be impossible to predict the time of excitation or recombination 
of a particular electron. However, using the methods of statistics it is 
possible to accurately predict the number of electrons in the conduction 
band or the number of holes in the valence band at a particular tempera- 
ture. This prediction is analogous to the accurate prediction of death 
rates of populations even though it may be impossible to predict the time 
of death of each individual. 

Fermi and Dirac used statistical methods to determine the probability 
of occupancy of any particular energy state as a function of temperature. 
This is analogous to the probability that a person will die within a year as a 
function of age. The number of persons in a given age group who will die 
within a year may then be determined by multiplying the total number of 
persons in the group by the probability function. In a similar fashion, the 
number of occupied states at a given energy level and temperature can be 
determined by multiplying the total number of states by the probability 
that the states are occupied. The Fermi-Dirac probability function is 

/(") = j , g (W-WF)lkT ^ 



34 Electronic Engineering 

where W = the energy level under consideration 

k = the Boltzmann constant, 1.38 x 10- 23 j/°K, which relates 
temperature to energy 
' T = the temperature in °K 

W F = a specific energy level known as the Fermi level 
When the temperature is absolute zero, it may be seen from Eq. 2. 14 that 
the probability is zero that an energy state of higher level than the Fermi 




_J_ 
l 

0.5 1.0 

f(w) — >- 
Probability function 

(a) 



Density of states 
(per unit energy) 

(b) 



Density of electrons 
(per unit energy) 

(O 



Nil- f (w)\—* 

Density of holes 
(per unit energy) 

(d) 



Fig. 2.6. An illustration of the method of calculating the densities of electrons per unit 
energy or holes per unit energy by taking the product of the probability function and the 
density of states per unit energy. 

level is occupied and is unity that an energy state having lower energy 
than the Fermi level is occupied. Therefore, a rule concerning the Fermi 
level may be stated. No electron can have energy higher than the Fermi 
level when the temperature is absolute zero. 

Again, Eq. 2.14 shows that the probability of an energy state being 
occupied is one half when the state is at the Fermi level and the temperature 
is above 0°K. This second rule is also useful in locating the Fermi level, as 
will be seen later. The Fermi-Dirac probability function is sketched for 
two different temperature values in Fig. 2.6a. 

In order to determine the number of occupied states, and hence the 
number of charge carriers in a given volume of crystal, the density of 
available energy states which have energies within a narrow energy band 



Semiconductors 35 

A W must be known as a function of energy. This, for example, is analogous 
to knowing the number of people in each age group as a function of age 
in the United States. The method of quantum mechanics provides the 
desired density of energy states having energies within the limits AW 
as a function of energy. First, Aldert van der Ziel has shown that the 
maximum possible density of electrons which can have velocities within the 
limits Av is given by the relationship 1 

A „ faOn.*)V Ap (215) 

h 3 

where AiV s is the number of available states per unit volume which can be 
filled by electrons having velocities within the limits At), m* is the effective 
mass of the electron, 2 and h is Planck's constant (6.625 x 10- 34 j/sec). 
Writing Eq. 2.15 in differential form, 

dN$ = Wme*fv 2 dv (2 16) 

The velocity v can be expressed in terms of the kinetic energy of the 
electron. Then substituting W k = mv 2 /2 and dW k = mo dv into Eq. 2.16 
we have 

s , 3 * 

When an electron receives sufficient energy to jump from the valence 
band to the conduction band, the potential energy of the electron is 
increased to the potential energy level of the conduction band, W c . 
The electron usually receives more than enough energy to jump the gap 
and become free. The excess energy is retained by the electron in the form 
of kinetic energy, W k . Therefore, the total energy of a conduction-band 
electron is W = W e + W k and the kinetic energy is W k = W — W c . 
Note that dW k — dW. Using these relationships, Eq. 2.17 becomes 

dN s = A C {W - W C Y A dW = A C N C {W) dW (2.18a) 

where 

A c = 8(2yMm e *) 3A lh 3 , 

and the density of states is 

1 See Solid State Physical Electronics, by Aldert van der Ziel, Prentice-Hall, Englewood 
Cliffs, New Jersey, Section 3.3, p. 46. 

2 Classical mechanics can be used to determine the behavior of free carriers in a 
crystal provided that an effective mass m* (which accounts for the periodic crystal 
forces) is used instead of the electronic rest mass m. (See Table 2.2.) 



36 Electronic Engineering 

It can be shown that the distribution of available states near the top 
of the valence band is the same as the distribution of states near the 
lower edge of the conduction band. In other words, Eq. 2.18a is applicable 
near the upper edge of the valence band when the term (W — W c ) is 
replaced by {W v - W) and the effective mass of the hole m h *, which is 
different than the effective mass of the electron, is used in the constant A. 

dN s = A V {W V - W)* dW = A v dN v (W) dW (2.18b) 

A sketch of the available states N, per unit energy as a function of energy 
Wis given in Fig. 2.66. 

TABLE 2.2 

Some Physical Constants and Some Fundamental 
Properties of Ge and Si 

Physical Constants 

Electron charge q, coulomb 
Electron rest mass m, kg 
Proton rest mass, kg 
Planck's constant h, joule-sec 
Boltzmann constant k, joule/°K 
Avogadro's number, molecules/kg-mole 
Permittivity of free space e„, farad/m 
Permeability of free space fi v , henry/m 
Speed of light, m/sec 
Wavelength range of visible light, A 



1.6 


X 


10 ~19 


9.11 


X 


10- 31 


1.67 


X 


io- 27 


6.62 


X 


10 ~34 


1.38 


X 


10 -23 


6.02 


X 


10 2 « 


8.854 


X 


10-12 


4tt 


X 


10~ 7 


3.00 


X 


10 s 


4000 


— 


7200 



Specific Properties Ge Si 



Atomic number 32 14 

Atomic weight 72.60 28.06 

Density (25 °C) kg/m 3 5.33 x 10 3 2.33 x 10 3 

Melting point, °C 936 1420 

Relative dielectric constant <r/e„ 16 12 

Effective mass ratio m*/m: Free electrons 0.55 1.1 

holes 0.37 0.59 

Intrinsic carrier density (300°K), n u 2.4 x 10 13 1.5 x 10 10 

Pi, cm" 3 

Gap energy ^.electron volts: 0°K 0.782 1.2 

300°K 0.67 1.1 

Carrier mobility (300°K), m 2 /v-sec /u n 0.39 0. 1 2 

f* P 0.19 0.05 

Diffusion constant (300°K) m 2 /sec D„ 98.8 x 10~ 4 33.8 x IO" 4 

D v 46.8 x 10-* 13 x 10- 4 



Semiconductors 37 

Now, the number of electrons, or occupied states, per unit volume which 
have energies within the band dW may be obtained from the product of the 
number of states dN s and the probability function f(W). 

dn =f(W)(dN s ) = A C (W- W c )^f{W) dW (2.19) 

where n is the number of electrons per unit volume. 

The number of electrons per unit volume (n) as a function of energy is 
shown in Fig. 2.6c when the temperature T = 7\. Since a hole is an 
unoccupied state, the density of holes per unit energy may be determined 
by subtracting the density of occupied states n, per unit energy, from the 
total density of states N„ per unit energy. Note that the probability 
function is symmetrical about the Fermi level. That is, the probability 
that states are filled at a given A W above the Fermi level is equal to the 
probability that states are empty [1 —f(W)] at the same AW below 
the Fermi level. The Fermi level is shown in Fig. 2.6a near the center 
of the forbidden band or gap. This position will be verified later. 

The total number of electrons per unit volume in the conduction band 
is obtained by integrating both sides of Eq. 2.19 over the entire conduction 
band. /•«, 

n= A C (W - W c ) A f(W) dW (2.20) 

JWc 

An examination of the probability function /( W), Eq. 2.14, suggests 
that a useful approximation may be made when W — W F is at least 
several times as large as k T. Then the exponential term is large in com- 
parison with one and ^^ _ e _ {w _ Wr)l1cT (2 21) 

For the normal room temperature of 300°K, kT = 4.14 x 10~ 21 joule. 
The normal forbidden band or gap energy is of the order of one electron 
volt = 1.6 x 10" 19 joule. The total gap energy is then about AOkT. 
Since the Fermi level is near the center of the gap, or forbidden band, as 
indicated in Fig. 2.6, there are no states closer than about 20kT to the Fermi 
level. Equation 2.21, which is known as the classical approximation, is 
highly accurate under these conditions. 

The number of occupied states, or electrons, per unit volume in the 
conduction band may now be determined. Using Eqs. 2.20 and 2.21 

n = f " A C {W - WS A e - KW : WF 'l* T dW (2.22) 

J We 

If a change in variable from ( W — W c ) to x is made, this integral may be 
recognized as a standard form. 3 With the aid of the integral table, 

n= dl ( kT )' A TT ,A e -Wc-Wr)l*T (2 23) 

2 
3 See Eq. 508 in B. C. Pierce and R. M. Foster, A Short Table of Integrals, Fourth 
Edition, Ginn and Co., New York, 1956. 



38 Electronic Engineering 

The previous technique may be used to determine the hole density in 
the valence band. The probability that states are not filled is 

1 - F W " t + e <wr-w»*T (2-24) 

PROB. 2.1. Verify Eq. 2.24. 

Since the energy level at the top of the valence band is several kT units 
below the Fermi level, 1 - F(W) ~ e -iw r -w)ikT^ Us j ng Eq 2 .\%b 

p = \ *A V (W V - W) 1A e - (w *- w ^ T dW (2.25) 

Notice the similarity between Eqs. 2.22 and 2.25. A serious question, 
however, may be raised concerning the validity of Eq. 2.25, as the density 
of states function is only valid near the top of the valence band but the 
integration is to be performed over the entire energy range below W v . 
This inconsistency will produce little error, however, because the proba- 
bility function 1 — F(W) approaches zero at an exponential rate, and at 
normal temperatures the product (1 - F(W))N S is negligibly small for 
all energy levels for which N s = A V {W V - 1V)X is not valid. The integral 
table may be used to integrate Eq. 2.25. Then 

p = 6s. ( k T) 3/i n A e - {w *- w ° ) i kT (2.26) 

Calculation of either p from Eq. 2.26 or n from Eq. 2.23 requires a 
knowledge of the Fermi level W F . W F , however, vanishes from the product 
pn as shown in Eq. 2.27: 

pn = 6i6s (^7)3^ e -<Wc-w„)i*T (2 2?) 

where W c — W v is the gap energy W g which is known. 

In the intrinsic material, the density of free electrons n t must be equal 
to the density of holes p t ; therefore, np = n i p i = «/ = p* and 

»* = Pi = i(A c A v 7rk 3 )' A T 3A e - lw ° ,2kT) = BT % e -< w ' l2kT) (2.28) 

The experimentally determined values of B (which give charge carriers 
per cubic meter) are : 

B = 1.76 x 10 22 for germanium 
B = 3.88 x 10 22 for silicon 

PROB. 2.2. Calculate the theoretical values of B for both silicon and germanium 
and compare them with the experimentally determined values. Why are the 
values of B different for germanium and silicon? 



Semiconductors 39 

A comparison between Eqs. 2.23 and 2.28 shows that the Fermi level 
would be at the center of the forbidden band or gap if A c were equal to 
A v . Actually, A c differs from A v because of the difference in the effective 
mass of the electron as compared with that of the hole in a crystal. 
Nevertheless, the Fermi level is very near the center of the gap in the 
intrinsic crystal. 

The gap energy is a function of temperature because the atomic spacing 
increases with temperature owing to thermal expansion. The gap energy 
can be determined from the following relationships : 

For silicon, W g = 1.2 - (2.8 x 10-*T) electron volts 

For germanium, W g = 0.782 - (3.9 x 10-*T) electron volts 

At 300°K the gap energy is approximately 1.1 electron volts for silicon 
and 0.67 electron volts for germanium. 

PROB. 2.3. Calculate the number of conduction electrons in 1 cc of intrinsic 

germanium at (a) 300°K, (b) 200°K, (c) 400°K. One electron volt = 1.6 x 10- 19 j. 

Answer: (a) 2.56 x 70 13 (b) 2.99 x JO 10 (c) 6.35 x 10™. 

PROB. 2.4. Calculate the number of conduction holes in 1 cc of intrinsic 

silicon at (a) 200°K, (b) 300°K, (c) 400°K. 

PROB. 2.5. Calculate the ratio of conduction electrons in germanium at room 

temperature (300°K) to the total number of atoms in the crystal. The density 

of Ge is 5.46 g/cm 3 . Its atomic weight is 72.6 grams, in which there are 6.024 x 

10 23 atoms (Avogadro's number). 

2.5 RECOMBINATION 

From the foregoing discussion, it is evident that free electrons and holes 
are being continually generated in a semiconductor crystal at normal 
temperatures. The number of charge carriers would increase indefinitely 
if the holes and electrons did not recombine. When the temperature of 
the crystal remains constant for a period of time, however, the rate of 
recombination is equal to the rate of generation and the carrier density 
remains essentially constant at the value predicted by statistical methods 
(Eq. 2.28). If the number of carriers were increased above the equilibrium 
value by some process, such as shining light on the crystal, the recombina- 
tion rate would also increase because the opportunities for recombination 
are proportional to the number of carriers. This is analogous to the 
marriage rate being proportional to the population. After removing the 
light, the recombination rate exceeds the generation rate and the carrier 
density decreases exponentially toward the equilibrium value. This process 
is expressed by the following relationship; considering negative carriers: 

n = n + An e~ t/T (2.29) 



40 Electronic Engineering 

where n is the number of negative carriers at equilibrium, An is the increase 
in carriers due to the light, and r is defined as the time required to reduce 
the excess carriers to 1/e times the number at the instant the light was 
removed at t = 0. t is given the name "lifetime" and compares with the 
time constant of an electrical RC circuit. The rate of change of carrier 
density 

*5 = -^V/' (2.30) 

dt t 



But from Eq. 2.29 
Then 



An e~ t/T = n-n (2.31) 

^ = 2° - 2 (2.32) 

dt T T 



The first term of Eq. 2.32 is the rate of generation and is a function of the 
temperature and the lifetime. The second term is the rate of recombination. 
It is proportional to the instantaneous carrier density and inversely 
proportional to the lifetime. When n is equal to w , the recombination 
rate is equal to the rate of generation and the density is in equilibrium. 

According to Dewitt and Rossoff, 4 experimental measurements of the 
lifetime of various crystals reveal that the lifetime of similar crystals varies 
widely. The purer crystals have longer lifetimes. In addition, the shape 
of the crystal has a pronounced influence on the lifetime. The larger the 
ratio of volume to surface area, the longer is the lifetime. For example, 
a cubic crystal has a longer lifetime than a thin or long crystal. The 
measured lifetimes are always much shorter than predicted values based 
on the assumption that the recombination results purely from an electron 
coming in contact with a hole. 

Investigators have postulated that the imperfections in the crystal act as 
a catalyst to speed up the recombinatior process. It is not known how 
these imperfections do this, but it is speculated that the imperfection some- 
how holds the carrier until an opposite type carrier comes along and 
recombination results. The probability of recombination would be 
increased by this mechanism. The imperfections are known as deathnium 
centers or traps. Actually the entire surface of the crystal is an extended 
imperfection. Therefore, a decrease in surface area for a given volume 
of crystal would reduce the number of traps and increase the lifetime of 
the carriers. 

4 The basic information for this discussion of lifetime and recombination was obtained 
from Transistor Electronics, by D. Dewitt and A. L. Rossoff; McGraw-Hill, New 
York, 1957. 



Semiconductors 41 

2.6 DRIFT AND CONDUCTIVITY 

The motion of the charge carriers would be random in the absence of an 
electric field. According to the particle theory of matter, each free electron 
would travel in a straight path until it had a collision. If the collision 
were elastic, the electron would proceed in another direction. By this 
process the motion would be completely random but there would be an 
average length of path between collisions. This average path is known as 
the mean free path. Also there would be a mean free time and an average 
particle velocity associated with this time. The holes would also move from 
atom to atom in a random fashion. The hole mean free path would 
normally be shorter than that of the electron. 

When an electric field is applied to the crystal, the carriers acquire a 
component of velocity in the direction of the field. Because of the increased 
kinetic energy of the carriers, they experience a greater energy loss at each 
collision. This energy loss accounts for the ohmic power loss in the 
conduction process. The carriers are deflected in random directions after 
each collision, but the component of velocity acquired in the direction of 
the field between collisions results in an average drift velocity v. The 
current density is equal to the product of the charge per unit volume and 
the drift velocity. Consequently, the current density due to electron drift 

/„ = nav„ (2.33) 

where v n is the average drift velocity of the electrons. 

The hole drift would also contribute to the current density. Therefore, 

J„=pqv P (2.34) 

where v p is the average drift velocity of the holes. 
The total current density 

J = J n + J P = q(nv n + pv p ) (2.35) 

Conductivity a is defined as the ratio of current density to the electric 
field intensity. 

a = j (2.36) 

Then, combining Eqs. 2.35 and 2.36 

a = a(nv n + pvj (23?) 

It is convenient to use the concept of charge mobility fx which is denned as 

P = 1, (2-38) 



42 

Then 



a = q(nfi n + pfi p ) 



Electronic Engineering 



(2.39) 



where fi n = the mobility of negative carriers 
fi P = the mobility of positive carriers 
The carrier mobilities for a material may be determined experimentally by 
the use of the "Hall effect" which will be discussed later. The mobilities 




Temperature 
Fig. 2.7. Conductivity of germanium and silicon as a function of temperature. 

of lightly doped germanium and silicon at 300°K as determined by 
M. B. Prince 5 are listed below. 





/*« 


<"» 


Germanium 


0.39 


0.19 m 2 /vsec 


Silicon 


0.12 


0.05 m a /vsec 



Carrier mobility decreases with increasing temperature because the 
increased vibrational energy increases the effective diameter of the atoms 
and thus reduces the mean free path of the carriers. The mobility tempera- 
ture coefficient is often neglected for temperatures near 300°K because 
it is dwarfed by the exponential variation of thermal generation with 
temperature. 

PROB. 2.6. Calculate the conductivity of pure germanium at room temperature 
(300°K). See Prob. 2.3. Answer: 2.22 mhos/meter. 

PROB. 2.7. Calculate the conductivity of pure silicon at 300°K. See Prob. 2.4. 
PROB. 2.8. What is the resistivity of (a) germanium and (Jb) silicon at 300°K? 
Answer: (a) 0.45 ohm-meter. (6) 2400 ohm-meter. 

Figure 2.7 illustrates the variation of conductivity of intrinsic germanium 
and silicon as a function of temperature. This variable conductivity 

* M. B. Prince, Phys. Rev., Vol. 92, No. 3 (1953). 



Semiconductors 43 

property is exploited in devices such as thermistors, which are used in 
applications such as thermal compensation of transistor circuits and 
automatic temperature control systems. In the thermistor, the nature of 
the variation of resistance over a given temperature range depends on 
the type of semiconductor material used, whereas the actual resistance at a 
given reference temperature depends on the size and shape of the device. 

PROB. 2.9. A rectangular thermistor is made of pure germanium. Its dimen- 
sions are 0.1 cm x 0.2 cm x 1.0 cm. The electrodes are attached to the ends 
of the bar. Determine the resistance of the thermistor at 300°K and 350°K. 
Assume that the carrier mobilities at 350°K are approximately the same as they 
are at 300°K. 



2.7 DOPED SEMICONDUCTORS 

A semiconductor crystal which is impure or imperfect may be called an 
extrinsic semiconductor. As previously discussed, imperfections cause 
traps or deathnium centers. It would be expected that impurities would 
alter the energy levels and change the conductivity of the crystal. When 
impurity atoms are intentionally added, the semiconductor is said to be 
doped. Very small amounts of impurity may have a remarkable effect on 
the behavior of the crystal. 

The conductivity of a semiconductor may be greatly increased if small 
amounts of certain impurities are introduced into the crystal. For example, 
if a few parts per million of elements such as arsenic or antimony, which 
have five valence electrons, are added to the semiconductor, the impurity 
atoms replace the germanium or silicon atoms in the crystal structure, as 
shown in Fig. 2.8. Four of the valence electrons will form covalent bonds 
with the neighboring atoms. The fifth electron will not fit into the lattice 
arrangement and, therefore, will be loosely bound to its nucleus. At 0°K 
the energy of this electron is only a few hundredths of an electron volt 
below the conduction band. Therefore, at normal temperatures the 
probability is very high that this electron will be a free electron in the 
conduction band. The electron is then said to be activated. Note that a 
negative carrier is produced without creating a hole or a positive carrier. 
A fixed positive charge remains in the lattice but no covalent bonds are 
broken. 

A semiconductor to which a pentavalent atom has been added is called 
«-type because the majority of charge carriers are negative. The penta- 
valent impurity atoms are called donors because they provide the extra 
free electrons. If there is only one donor for each million intrinsic atoms, 
the number of free electrons and hence the conductivity may be increased 
many-fold. In n-type material, the holes produced by thermal generation 



44 



Electronic Engineering 




Loosely 

bound 

electron 



Fig. 2.8. Crystal lattice with a donor impurity. 




Fig. 2.9. Crystal lattice with an acceptor impurity. 



Semiconductors 45 

readily recombine because of the abundance of free electrons. Therefore, 
conduction is due almost entirely to free electrons. In this case, the elec- 
trons are called majority carriers and the holes are called minority carriers. 
Consequently, the conductivity is 

a ~ q nf i n (2.40) 

Each donor atom provides an allowed energy state in the forbidden band 
(of the intrinsic material) at an energy level of a few hundredths of an 
electron volt below the conduction band, as shown in Fig. 2.10a. This 
level is known as the donor level. 

PROB. 2.10. Assuming that all the donor atoms provide a free electron, calcu- 
late the conductivity at 300°K of //-type germanium which has one donor atom 
per million germanium atoms. Compare the conductivity with that of pure 
germanium (Prob. 2.6). 

If a small quantity of trivalent impurity is added to an intrinsic semi- 
conductor, each impurity atom fits into the crystal structure but lacks one 
of the electrons needed to complete the covalent bonds. This situation is 
illustrated in Fig. 2.9. Each impurity atom is called an acceptor because 
it provides a hole or positive carrier. At normal temperatures these holes 
move freely through the material and become charge carriers because very 
little energy is required to move a valence electron from a neighboring atom 
into the hole provided by an acceptor atom. 

Each acceptor atom provides an allowed energy state in the intrinsic 
forbidden band at an energy level a few hundredths of an electron volt 
above the valence band as shown in Fig. 2.10a. This level is known as 
the acceptor level. 

A crystal which is doped with acceptor atoms such as indium or gallium 
is known as p-type because conduction is primarily by holes. In />-type 
material, the free electrons which are thermally generated in the intrinsic 
material quickly recombine because of the abundance of holes. Therefore, 
the conductivity of />-type material is 

o ~ q Pf i p (2.41) 

In the event that both donors and acceptors were added to an intrinsic 
semiconductor, it would be expected that the donor electrons would com- 
bine with the acceptor holes. The difference between the donor and 
acceptor concentrations should determine the effect of the doping. Equal 
concentrations should provide characteristics similar to the intrinsic 
material. Experimental evidence verifies that these assumptions are 
essentially correct. 



46 



Electronic Engineering 



The Fermi-Dirac distribution function is also applicable to the extrinsic 
or doped semiconductor. Determining the location of the Fermi level is 
usually the major problem in the application of this function. The Fermi 
level was found to be near the center of the forbidden band in the intrinsic 
semiconductor; the knowledge that the density of free electrons is equal 



w c 
w d 






II II lllllll 



3DBEDB 



EI El 
-i-i-i-i-M-i-i-i-n 




(a) 



(b) 



Fig. 2.10. Energy states of a "doped" semiconductor. 



to the density of holes was used to make this determination. Similarly, 
the doping concentrations and the degree of activation of the impurity 
atoms must be known in the extrinsic semiconductor in order to locate 
the Fermi level. The degree of activation, and thus the Fermi level, is a 
function of temperature. Before we attempt to locate the Fermi level in 
a specific example, let us review the characteristics of the Fermi-Dirac 
distribution function. 

1. The probability that states with energies above the Fermi level are 
filled is always less than one half. 

2. The probability that states with energies below the Fermi level are 
filled is always more than one half. 

3. At 0°K all states with energies above the Fermi level are empty and 
all states with energies below the Fermi level are filled. 

Now let us consider a silicon crystal which has been doped with donor 
atoms. The foregoing rules clearly show that the Fermi level must be 
above the donor level at all temperatures below T ly which is the temperature 



Semiconductors 



47 



required to activate one half of the donors (see Fig. 2.106). For tempera- 
tures above 7\, the Fermi level must drop below the donor level because 
less than half of the donor states are filled. 7\ is of the order of 50°K. 
At temperatures well above T x , almost all the donors are activated and the 
negative carrier density is approximately equal to the donor density 
because the carriers generated from the intrinsic atoms are normally 
negligible. Whenever the carrier density is known, the Fermi level may 
be determined at a given temperature by the use of Eq. 2.23 because the 
Fermi level with reference to W c is the only unknown in the equation. 

PROB. 2.11. A silicon crystal is doped with 10 21 donors per cubic meter. 
Find the Fermi level with respect to the bottom of the conduction band at 
300°K. Assume the constant AJ&ttMjI to be approximately equal to the 
constant B (Eq. 2.28). Answer: 0.316 v below W c . 

PROB. 2.12. Determine the probability that the donor states are filled in the 
doped crystal of Prob. 2.1 1 Assume that the donor level is 0.02 electron volts 
below the conduction band. 

As the temperature is raised to such a high value that the carriers 
generated by broken covalent bonds in the intrinsic material have a 
density comparable with the donor density, the Fermi level must approach 
the center of the forbidden band because the hole density will then be 
comparable with the electron density and this situation can occur only 
when the Fermi level is near the center of the gap. The Fermi level is 
sketched as a function of temperature in Fig. 2.10Z>. A sketch of charge 
carriers as a function of temperature is shown in Fig. 2.11. 



(N d -N a ) 




^L 




Ti T a 



T 2 



Fig. 2.11. A typical plot of charge carriers as a function of temperature for n-type 
crystal. 



48 Electronic Engineering 

PROB. 2.13. Sketch the Fermi level as. a function of temperature for a/?-doped 
crystal. 

It was previously shown by Eq. 2.27 that the product of positive and 
negative charge carriers pn is independent of the Fermi level, assuming a 
condition of thermal equilibrium. Therefore Eq. 2.27 is valid for doped 
as well as intrinsic crystals. In the intrinsic material, since p t = n t , we 
can write 

pn = n? (2.42) 

where n { is the number of negative carriers in thermal equilibrium. In the 
doped crystal, the Fermi level is shifted away from the center of the gap 
and/? is not equal to n. However, the product pn is a function of the gap 
energy and temperature only and is, therefore, the same as in the intrinsic 
crystal. Then 

p = ^ (2.43) 

n 

Equation 2.43 is useful in determining the number of minority carriers 
per unit volume when the number of majority carriers and intrinsic carriers 
are either known or can be calculated easily. Inspection of Eq. 2.42 
reveals that in the doped semiconductor the minority carriers are reduced 
from the number in the intrinsic material by the same ratio as the increase 
in majority carriers, which results from the doping. For example, consider 
an n-type crystal which has one hundred times as many negative carriers 
per unit volume as the intrinsic crystal would have at the same temperature. 
Thep-type carriers would only be 0.01 as numerous as those in the intrinsic 
material. Physically, the reduction in holes results from the increased 
opportunities for recombination offered by the activated donor electrons. 
Mathematically, the increased probability that the holes are filled results 
from the elevated position of the Fermi level, compared with its central 
position in the intrinsic case. The end result is that the conductivity of 
either n- or ^-type semiconductors is almost entirely due to the majority 
carriers. 

2.8 THE HALL EFFECT 

The phenomenon known as the Hall effect is significant in the discussion 
of semiconductor theory for two reasons. First, the Hall effect provides a 
method of determining the electron and hole mobilities for a given material. 
Second, the Hall effect demonstrates that conduction by holes is a different 
process than conduction by electrons. Frequently, the beginning student 
is reluctant to accept the concept of conduction by holes. 



Semiconductors 



49 



Figure 2.12 illustrates the Hall effect. A current is passed through a 
conductor or semiconductor as indicated. The current flows because of an 
axial, or x, component of electric field. A magnetic field is arranged so 
that the magnetic flux in the material is normal to the direction of current 
flow. Assume that the material is a p-type crystal and that the magnetic 
flux lines are coming out of the paper. The holes would then experience a 
downward force which would cause them to crowd together toward the 
bottom of the crystal as indicated by the closeness of the line spacing in the 



Hole current Equipotential surfaces 





Fig. 2.12. The Hall effect. 

figure. This uneven distribution of charge, would produce a transverse, or 
y, component of electric field. Since there can be no appreciable transverse 
current flow in the crystal, the upward electrostatic force on each charge 
must be equal to the downward magnetic force. The resultant electric field 
in the crystal would be the vector sum of the axial and transverse compo- 
nents, as indicated by the vector diagram of Fig. 2.12. Then equating the 
transverse electric and magnetic forces, 

F =BIl = pqA\S v (2.44) 

where / = the length of the crystal 
A = cross-sectional area 
pq = charge density due to the holes 
The axial component of electric field, using Eq. 2.36, is 

S s = - (2.45) 



From Eq. 2.44 



Then 



£„ = 



BI 

pqA 



6 ,, 



BJ 

PI 



tan = — ' = 



Bo 

pq 



(2.46) 



(2.47) 



50 



Electronic Engineering 



Using the relationship a = pqp v (Eq. 2.41) 

tan = B/i P (2.48) 

The flux density B can be calculated or measured. The angle 8 may be 
determined from careful measurements to find equipotential points on the 
face of the crystal. After a line is drawn through the equipotential points, 
the angle 6 is measured between the equipotential line and the transverse 
line. The hole mobility for the type crystal used may then be calculated. 

/>, = *¥ (2-49) 

B 

An «-type crystal could have been used for the preceding example 
instead of a/>-type. The force exerted by the magnetic field on the moving 
charges would still be downward and the electrons would be crowded 
toward the bottom of the crystal. Therefore, the transverse component of 
electric field would also be downward and the equipotential surfaces would 
slope the opposite direction from the transverse surface, in comparison 
with the />-type crystal. Thus, the type of conductivity may be identified. 

MKS units were used in the preceding development. If the magnetic 
flux density is given in webers/m 2 , the mobility will be obtained in m 2 /v-sec. 

PROB. 2.14. A Hall effect experiment is arranged as shown in Fig. 2.12. If 
the magnetic flux density B is 0.1 webers/m 2 and the angle which the equi- 
potential surface makes with the horizontal x direction is 80°, determine the 
mobility and the type of majority carriers. 

Actually the carrier mobility decreases as the temperature increases, 
because the carrier mean free path decreases. The mean free path decreases 

because the vibration amplitude of each 
atom about its equilibrium position increases. 
Therefore, the effective collision diameter of 
each atom increases with temperature. For 
this reason the resistance of a conductor 
increases with temperature. 

PROB. 2.15. Let us assume that there exists an 
element for which the portion of filled electron 
states is not given by the Fermi-Dirac distribu- 
tion but rather the function shown in Fig. 2.13. 
150 / T_ _ _ \ filled states 
"T\300 / states 




That is, / = 



Fig. 2.13. Configuration 
Prob. 2.15. 



where tV 1 > W > W 2 and/ = for W > W x 
and/ = 1 for W < W 2 . 

It has also been determined that 

_ available states 
N{W) = A(W - W. e ) = 



Semiconductors 51 

rather than A C (W - W C )X, where A = 10 20 and Wis in electron volts, for this 
particular element. (This expression is valid only in the lower part of the 
conduction band.) 

T , T 

Let JV C = -6 ev, W v = -8 ev, W x 7 + — , W 2 = -7 - — . 

(a) How many available states N are there with an energy of —5.5 ev in a 
cubic centimeter? 

(6) How many available states are there with energies between —6 and —5.5 
evinacm 3 ? 

(c) What is the maximum temperature in degrees Kelvin for which there will 
be no electrons in the conduction band ? 

(d) Does any such temperature exist in the case of the Fermi-Dirac distribu- 
tion? Why? 

(e) What is the most probable number of electrons per cm 3 with an energy 
between -5.75 ev and -5.74 ev if T = 450°K? 

(/) What is the maximum number of electrons per cm 3 which can be between 
the energy levels of —5.75 ev and —5.74 ev? 

(g) If T = 450°K, how many electrons will there be in the conduction band 
for 1 cm 3 of the material? 

PROB. 2.16. A large bar of germanium having the dimension shown in Fig. 
2.14 is used for the following problems. 




5 cm ■ 



1 




Uniform 
connection 

Fig. 2.14. Configuration for Prob. 2.16. 

(a) What is the resistance of the bar at 400°K if there are no impurities in the 
germanium? Neglect edge effects and assume that the bar is enclosed in a light- 
proof box. Also assume that the mobility does not change with temperature. 

(b) What changes will take place in the resistance if a bright light (which 
generates no heat) is suddenly turned on in such a way that all side surfaces are 
illuminated for one minute, then the light is turned off? 

(c) Find the resistance at 400°K (no light) if arsenic atoms (pentavalent) are 
added in the ratio of 0.5 parts per million. 

(d) Without calculating the new values of resistance, discuss the effects of 
reducing the temperature from 400°K to 300°K for the intrinsic germanium bar 
of part a and the doped bar of part c. 



3 



Diodes 



A diode is an electronic device that readily passes current in one direction 
but does not pass appreciable current in the opposite direction. Such a 
device is formed when a piece of w-type semiconductor is connected to a 
piece of /"-type semiconductor. In actual production, a single crystal of 
semiconductor is formed with half of the crystal doped with acceptor 
impurities and the other half of the crystal doped with donor impurities. 
Other types of diodes are constructed by placing an electron emitter 
(cathode) and an electron collector (plate) in a vacuum or in a controlled 
gaseous atmosphere. The properties of the various types of diodes are 
considered in this chapter. 

3.1 THE p-n JUNCTION 

Figure 3.1 illustrates the action of the p-n junction. Assume two 
pieces of semiconductor exist (Fig. 3.1a). The p material contains an 
extremely small percentage (in the order of 10~ 4 %) of atoms with a 
valence of +3. These acceptor atoms are represented in Fig. 3.1a as the 
circles with the negative signs. Associated with each acceptor atom is a 
hole, which is represented as a positive sign. An even smaller percentage 
of intrinsic atoms have lost electrons due to thermal agitation (or photo 

52 



Diodes 



53 



p 


L±J L^J 

eeee ee e 
+++++++ 

0996969 
+ + + + + + + 



CE — S 

©©© © © © ® 

©©©©©©© 



(a) 



ETUI 

e©e© ©e© 
++++++ 

©ee©©©© 
++++++ 



LtJ LJ 

©©©©©©© 

©©©©©©© 



(b) 



+ 






■> 











v. 


>> 






- 




L 







(0 



S 




Fig. 3.1. The p-n junction, (a) Representation of p- and n-type material; (b) represen- 
tation of p-n junction ; (c) charge distribution of the p-n junction; (d) potential distribu- 
tion of the p-n junction. 

emission). These atoms are represented by the positive sign in the square. 
The free electrons released by this action are represented by the negative 
sign in the square, in the n material of Fig. 3.1a, the +5 valence atoms 
are represented by the circles with the positive sign. The free electrons 
due to these donor atoms are shown as negative signs. As in the p 
material, the intrinsic atoms form a few electron-hole pairs. Again, these 



54 Electronic Engineering 

electrons are represented by a negative sign in a square and the holes are 
represented by a positive sign in a square. It is important to note that both 
the piece of p material and the piece of n material are electrically neutral. 

A charge redistribution occurs when the two pieces of semiconductor 
materials are connected. In this condition, as shown in Fig. 2>.\b, some of 
the free electrons from the n material have migrated into the p material 
and have recombined with the extra holes. Also, some of the holes from 
the p material have migrated into the n material and have recombined 
witn free electrons. As a net result, the p material has assumed a net 
negative charge and the n material has assumed a net positive charge. 
This charge distribution is shown in Fig. 3.1c. As a result of the charge 
distribution, a potential difference exists across the junction (Fig. 3. Id). 

The process by which the charges cross the junction is known as diffusion. 
This process may be visualized if only the action of the electrons from the 
n piece of material is considered. When the two pieces of material are 
joined, there is a concentration of electrons in the n material but essentially 
none in the p material. The random motion of the electrons will allow 
some electrons to pass from the n to the p region. Because there are 
fewer electrons in the p region than in the n region, fewer electrons will 
tend to pass from the p to the n region. If no other forces exist, the 
diffusion process will continue until the concentration of electrons is 
uniform throughout the material. This process is the same process which 
occurs when two containers of dissimilar pure gases are joined. Eventually 
both containers will contain a uniform mixture of both gases. However, 
because of the charges on the electrons, these electrons are attracted toward 
the positively charged n region. Similarly, the holes are attracted toward 
the negatively charged p region. Hence, an electric field is established 
(by the diffusion process), which inhibits the diffusion process. 

A crude analysis of the diffusion process can be made. In this analysis, 
assume an electron distribution exists as shown in Fig. 3.2. The charge on 
the electrons will be ignored in this analysis but will be considered later. 
Because of the random distribution of electron directions, half of the 
electrons will have a component of velocity to the left and half will have a 
component of velocity to the right. Now, assume that all the electrons 
contained within the slice of crystal between x and (x + /,) are able to 
move a distance / in the x direction in the time i. The term / is the mean 
free path of the electron in the x direction and t is the time required to 
travel this distance. Then, the total number of electrons crossing the line 
x of Fig. 3.2 from the right is one half the total electrons contained 
between a- and (.r + /). But, the total number of electrons in a unit cross- 
sectional area is equal to the electron density per unit length multiplied by 
the total length. In this process, the total number of electrons per unit 



Diodes 



55 



cross section is equal to the area under the curve of Fig. 3.2 between the 
points x and {x + /"). The total number of electrons per unit cross section 
is therefore approximately 

The first term In on the right of the equation is area 1 of Fig. 3.2; the 




Fig. 3.2. A plot of electron distribution for use in deriving the diffusion equations. 

second term is approximately equal to area 2 of the same figure. Area 2 

dn - 
is a triangle and hence has an area of \l An where An is given as — /. 

The total number of carriers per unit cross- sectional area which cross x 
from the right in time i is, therefore, 



NR = U nl+ idj 1 . }2 \ 

R 2\ 2dx I 



(3.2) 



By similar reasoning, the total number of carriers per unit cross-sectional 
area which cross x from the left in time i is 



L 2\ 2dx J 



(3.3) 



The net flow of electrons per square meter across the surface at x in time 
f is 



N m = N, 



N T . = i/ 2 



T>dn 



dx 



(3.4) 



56 Electronic Engineering 

The total flow per square meter per second is 

1 I 2 nd 

N * = Z-— (3.5) 

2 t ax 
or 

lv n dn 
N t = ~ n - (3.6) 

2 dx 

where v is the mean thermal velocity of the electrons and is equal to I/f. 
Since each carrier has a charge q, the total current flow per square 
meter across the plane x is equal to 

2 az 

A similar equation for current flow due to the diffusion of holes can be 
derived. In this case, 

-a l f d f (3.8) 

2 dx 

where J P = the current density in amps/m 2 (due to hole movement) 
/„ = the mean free path of the holes in m 
Vj, = the mean thermal velocity of the holes in m/sec 
p is the concentration of holes/m 3 

The term - of Eqs. 3.7 and 3.8 is called the diffusion constant D. 
Consequently, 

J n =qD n ^ (3.9) 

dx 
and 

J p =- q D v i? (3.10) 

dx 

The diffusion constant is a function of the type of material and the 
temperature as well as the type of carrier. The values for lightly doped 
germanium and silicon at 300°K are given in Table 3.1. 

TABLE 3.1 
Values for Germanium and Silicon at 300 K 

Ge Si 



D n 98.8 x 10" 4 33.8 x 10- 4 m 2 /sec 
D„ 46.8 x 10" 4 13.0 x 10- 4 m 2 /sec 



Diodes 57 

In the actual n-p junction, as stated previously, both types of semi- 
conductive materials are neutral before the junction is made. After the 
junction is made, the n material loses electrons due to diffusion and gains 

holes. 

However, as the electrons enter the /7-region, recombination with the 
numerous holes occurs. In fact, the hole population next to the junction 
is said to be "depleted," since practically all these holes are filled with 
electrons. The original electrically neutral nature of this region is replaced 
with negative ions wherever a +3 valence atom exists. Similarly, any 
holes which "travel" (or are injected) into the n-material are immediately 
removed by recombination with the free electrons near the junction. In 
the n-material, the free electrons are depleted and positive ions occur 
wherever a +5 valence atom exists. As a result of this action (the p- 
material becomes negative and the n-material becomes positive), an 
electric field is established at the junction. In the steady-state condition, 
this field is just strong enough to inhibit the diffusion action of the electrons 
from the n-material and the holes from the /^-material. 

The free electrons in the n material and the holes in the p material are 
known as majority carriers. But, because of thermal energy, photons, etc., 
some free electrons are produced in the p material and some holes are 
produced in the n material (as indicated by the signs in the squares in 
Fig. 3.1). Some of these carriers diffuse to the junction and are swept 
across the junction by the electric field. As mentioned in Chapter 2, 
these carriers are known as minority carriers. Since the potential hill at 
the junction is caused by the diffusion of the majority carriers, the minority 
carriers which are swept across the junction tend to reduce the height of 
the potential hill. As a result, an equilibrium condition is reached in 
which the flow of minority carriers is equal to the flow of majority carriers. 
The current resulting from the diffusion of the majority carriers is known 
as the injection current and the current resulting from the minority carriers 
is known as the saturation current. 

As previously mentioned, the charge distribution of Fig. 3.1c causes a 
potential difference across the p-n junction (Fig. 3.1rf). This potential 
difference is a few tenths of a volt at room temperature. It seems possible 
that a current would flow in an external circuit if a conductor were 
connected to the open ends of the p-n combination since they are at 
different potentials. This supposition is not true, because the contact 
difference of potential at the junctions of the crystal and the external 
conductor causes the total emf around the closed circuit to be zero, and 
no current will flow. For example, if a conductor is brought into contact 
with an n crystal, the net drift of electrons will be from the crystal to the 
conductor until the crystal becomes positive with respect to the conductor 



" Electronic Engineering 

and equilibrium is reached. 1 Of course, the opposite is true if the conductor 
is connected to a p-type crystal. 

The directions of the injection current I z and the saturation current I s 
are shown in Fig. 3.3b. In addition, the lengths of the current arrows 
indicate the relative magnitude of these currents. The net current flow in 
this case is zero, I z = I s . The majority carriers must have enough kinetic 
energy to "climb" the potential "hill" whereas all minority carriers just 
"slide down" this potential hill with no kinetic energy required. In fact, 
the minority carriers gain kinetic energy in traversing the junction. 

If an external battery is connected with the positive terminal to the 
n-type material and the negative terminal to the p material, the potential 
difference across the junction will increase as shown in Fig. 3.3c. The 
height of the potential "hill" is increased by the amount of the external 
battery voltage. Then only those majority carriers which have a very large 
amount of kinetic energy can "climb" the potential "hill." As a conse- 
quence, the injection current I r is greatly reduced as shown in Fig. 3.3c. 
Since the minority carriers still require no energy, the current I s remains 
the same as in Fig. 3.3b. The external current flow is, therefore, mainly 
owing to the thermally generated minority carriers of the semiconductor. 
This current I s — I T is known as the reverse current of the diode. 

If the external battery is connected with the negative terminal to the 
n-type material and the positive terminal to the p material, the potential 
difference across the junction is decreased as shown in Fig. 3.3d. The 
potential "hill" is reduced by the magnitude of the external battery voltage, 
neglecting the IR drop in the crystals. As a result, a large number of the 
majority carriers are able to cross the junction. Therefore, the injection 
current I 7 is greatly increased as shown in Fig. 3.3d. Since the minority 
carriers can still traverse the junction with no loss of energy (these carriers 
still gain kinetic energy but not as much as in Fig. 3.3b and Fig. 3.3c), 
the current I s does not change. The forward current of the diode I t — I s is, 
therefore, quite large. In fact, because of the large number of majority 
carriers as compared to the minority carriers, the forward current is 
usually thousands of times larger than the reverse current. 

PROB. 3.1. A concentration of 10 13 electrons/cm 3 exists in germanium at point 
x and to the right of point x. The distribution decreases linearly to the left until 
1 cm from x no electrons are present, (a) At the instant this condition exists, 
what is the rate at which diffusion electrons are passing a plane 0.01 cm to the 
left of point xl (Neglect the effects of charge on the electrons to find diffusion 
rate.) What current density does this electron flow represent ? (b) Repeat part a 
for a plane J cm to the left of point x. (c) Repeat part a for a plane 1 .01 cm to the 
right of point x. Assume the mean free path of an electron is less than 0.01 cm. 
1 This establishment of a potential difference between a doped semiconductor and a 
conducting wire is used in the construction of point contact diodes: 



Diodes 



59 



(a) 




- 



- 




*s h 






+ 

o 

> 


(c) 






c 




T 


f>W 






r 




r 


c 




o 

Q_ 










Is 


I, 




- 




(d) 







Fig. 3.3. The effect of an external voltage on the potential distribution of a p-n junction, 
(a) The. p-n junction; (6) potential distribution with no external battery; (c) potential 
distribution with a reverse bias external battery ; (d) potential distribution with a forward 
bias external battery. 



60 



Electronic Engineering 



3.2 THE DIODE EQUATION 

When p and n materials are joined together, the Fermi levels of each 
must be at the same absolute level in order to provide thermodynamic 
equilibrium. The proof of this statement is beyond the scope of this 
book. Nevertheless, if the Fermi levels were not the same, a voltage 
could be measured between the open ends of the material, and the junction 
diode would act as a battery. 2 However, since no energy source is present 
in the material, it cannot generate an emf. Figure 3.4 is a diagrammatic 
sketch of the energy levels of the p-n combination. In the figure, W cp is 





P 




P 


n 




n 




Conduction band 


w cp 


Conduction band 

} 


— T - 
1 qVho 

\ I 




onduction band 


Fermi 


Wf 

w » 

"P Valence band \ 
W„„ 






level 






Valence band 


\ 





Fermi 
level 



Fig. 3.4. Energy levels of a p-n junction. 

the lowest conduction-band energy level in the p material and W cn is the 
corresponding level in the n. In the n material, the Fermi level is near the 
donor level and, therefore, just below the conduction band as discussed in 
Chapter 2. In the/? material, the Fermi level is near the acceptor level, just 
above the valence band. As the junction is formed, the diffusion of carriers 
occurs until the Fermi levels become aligned. The difference between the 
two conduction-level energies then becomes q V h0 , where V h0 is the potential 
difference between the conduction levels in volts. V h0 was called the 
potential hill in the previous section, but is also known as the barrier 
potential. 

Since electron energies are shown in Fig. 3.4 and the potential energy 
of positive carriers or holes is shown in Fig. 3.1, the potential hill in 
Fig. 3.4 is inverted as compared with the one in Fig. 3.1. 

Using the technique developed in Chapter 2, we may determine the 

2 A p-n junction can be forced to act as a generator if external energy is supplied in 
the form of heat or light, in which cases the Fermi level is displaced. These devices 
are known as thermoelectric and photovoltaic generators, respectively. 



61 
Diodes 

negative carrier density for each type of material. In the n material (see 
Eq. 2.23), 

n = A ' T 3 A e -lWcn-Wf)lkT _ (3.11) 

All the constants of Eq. 2.23 have been absorbed by A' in Eq. 3.11. 

Similarly 

„ = A ' T ^ e -^Wcv-w,)/kT (3.12) 

where n p is the negative carrier density in thep material. The ratio of Eqs. 
3.11 and 3.12 is, 

Us = g-Wcr-WeJIkt (3.13) 



n. 



But the energy difference, W cv - W en = qV h0 . Then 

n v = n n e-" v "^ T (3-14) 

Equation 3.14 shows that the ratio of negative carriers in the two regions is 
dependent only on the barrier potential between these regions (for a given 
temperature). When an external voltage is applied to the diode, the barrier 
potential becomes V h and 

„„ = n n e-" v ^ T (3-15) 

These electrons, n„ are the ones which have sufficient energy to cross the 
potential barrier and, in so doing, become part of the injection-current 
density, 7 7 . The remaining part of the injection-current density results from 
the holes which have sufficient energy to cross in the opposite direction. 
From the discussion in Chapter 2 and the foregoing development, it is 
apparent that 

The injection-current density is proportional to the number of majority 
carriers which pass through the potential barrier V h . Then 

J t = B\n n + p,*-*™* (3-17) 

But n H andp„ are the majority-carrier densities in their respective materials 
and are essentially equal to the densities of donors and acceptors. There- 
fore, they may be included in a new constant B. 
Then 

J t = Be-" , ' ft/fcT (3-18) 

When there is no external connection, the injection-current density must 
be equal in magnitude but opposite in direction to the saturation-current 
density, as previously observed. Then 

J f ,= -Be- vrulkT (3-19) 



62 Electronic Engineering 

where F ft0 is the height of the potential barrier when no external voltage 
is applied as previously indicated. Solving for B in terms of J s , 

. „ , lo B=-J 8 e?>™*T (3.20) 

and, using Eq. 3.18, 

J z = -J se oW h o-V h )llcT ^2^ 

but the difference between V h0 and V h must be due to an external voltage, 
V. Then 

J z = -J s e° v >« T (3.22) 

where V is part of the externally applied voltage. This voltage V differs 
from the externally applied voltage by the IR drop in the doped crystal. 3 
The injection current is obtained by multiplying both sides of Eq. 3.22 by 
the cross-sectional area at the junction. Hence, 

h = -I s e« v '« T (3.23) 

The algebraic sum of the injection and saturation currents must be the 
current in the external circuit. 

/ = -I s e« v <* T + I S = -I s {e" v '^ - 1) (3.24) 

At room temperature (300°K) Eq. 3.24 reduces to 

/ = -i s(e ™v _ 1} (3 25) 

If V is about 0.1 volt or more negative, / is approximately equal to I s . 
If V is 0.1 volt or more positive, the exponential term of Eq. 3.25 is large 
in comparison with unity, and / is approximately equal to the injection 
current. 

The maximum rated current for a given diode is determined by the heat 
dissipation qualities of the mounting system and the cross-sectional area 
of the diode. In addition, the type of material used in the diode has a 
bearing on the maximum current rating. 

PROB. 3.2. When 1 v reverse bias is applied to a junction diode, 1 fia. of 
current flows. Calculate the approximate forward current when 0.256 v forward 
bias is applied to the junction. If the diode has 10 a internal resistance, what is 
the magnitude of external voltage required in order to obtain 0.1 v across the 
junction ? Answer v = 0.476 volts. 

PROB. 3.3. Plot a curve of /vs. Kas given by Eq. 3.25 if / s is lO" 6 amp. Plot 
values from —5 v to +0.3 v. 

3 This is a point which should be stressed. Quite often an external battery is applied 
to the diode and the current through the diode is measured. Since these values do not 
agree with values obtained from Eq. 3.23 (with K bal used for V) the reader may assume 
Eq. 3.23 is not valid. However, remember Kis not equal to the battery voltage K, , but 
will always be less. J 5 tet 



Diodes 



63 



3.3 SEMICONDUCTOR DIODE CHARACTERISTIC CURVES 

A sketch of current vs. voltage as given by Eq. 3.24 is shown as the solid 
line in Fig. 3.5. The plot of current vs. voltage for an actual diode is 
similar to the expected curve, but considerable departure exists for large 
values of both negative and positive bias voltage. This departure is shown 
by the dashed line in Fig. 3.5. The departure in the reverse bias region from 
a to b is partially due to leakage along the surface of the junction. Other 



Theoretical 




Fig. 3.5. Characteristic curve of a p-n junction diode. 



leakage paths exist through the mounting material which protects the 
junction. As a result, this departure is known as the leakage component. 

At point c of Fig. 3.5, a new effect known as the breakdown of the crystal 
is noted. In this "breakdown" region, high currents may be passed and 
they are limited only by the resistance in the external circuit. These high 
currents may generate enough heat to destroy the junction. Consequently, 
most diodes have a maximum reverse voltage rating. This maximum 
reverse voltage depends somewhat on the temperature of the diode. The 
"breakdown" region of the diode, as well as the region from b to c, is 
investigated in more detail in Section 3.6. 

The departure from the theoretical curve in the forward bias region is 
caused by the IR drop in the doped crystal as previously mentioned. 



°4 Electronic Engineering 

3.4 TEMPERATURE AND SEMICONDUCTOR MATERIAL 
EFFECT ON THE CHARACTERISTIC CURVE 

Equation 3.24 shows that the diode current is a function of temperature, 
which appears in the denominator of the exponent. However, the varia- 
tion of the saturation current with temperature is much greater than the 
variation of the exponential term. The saturation current is proportional 
to the number of minority carriers which are swept across the p-n junction 
as previously discussed and the number of minority carriers swept across 
the junction is proportional to the minority-carrier density in the material 
under consideration. Using the relationship pn = n ( 2 (see Eq. 2.42), 

„ — n ? ~ n * 2 (1 ->r \ 

Pn = — — — (3.26a) 

and 



n, 2 rii 

n„ = — - ~ — 



(3.26b) 
Then '» "' 

J s ~ C(p n + n p ) = C m 2 (— + — ) = C n? (3.27) 

\N a Nj 

where C and C" are constants. Using Eq. 2.28, we see that 

pn = n, 2 = B 2 T 3 e~ lw ' /kT) (3.28) 

The rate of change of J s with temperature may be obtained by differentiating 
J s with respect to temperature. 

The fractional increase of J s per °K can be determined at a given tempera- 
ture T by dividing Eq. 3.29 by / s . Then 

1 (dJ s \ 3 W„ 

7SE) = T + l£ (3 - 30) 

For germanium at 300°K, the fractional increase of J s per °K is 

For silicon at 300°K, the fractional increase per °K is 

Ite) «-L + 3 9 (lL) = o.l6 (3.32) 

J s \dTl 3oo°k 300 \300/ 



Diodes 



65 



Therefore, the saturation current in germanium near 300°K increases 
approximately 10 per cent for each degree K and doubles for each 10°K 
increase in temperature, and the saturation current in silicon doubles for 
approximately each 6°K increase. 

The foregoing derivation is valid for the semiconductor in the extrinsic 
range (or the range where the material behaves as n or p material). Germa- 
nium begins to be useless as a diode material at 100°C. However, silicon 



Tens of volts 




Fig. 3.6. The effect of increased temperature on the characteristic curve of ap-n junction 
diode. 

does not generally begin to degrade too badly until 200°C or higher. 
Other materials such as silicon carbide* are usable to temperatures of 

500°C or higher. 

The effect of increased temperature on the characteristic curve of ap-n 
junction diode is shown in Fig. 3.6. 

PROB. 3.4. (a) If a germanium diode has a saturation current I of 10 ^a at 
room temperature (300°K), what will the saturation current be at 400°K? 
(b) A silicon diode has a saturation current I of 0.01 //a at room temperature. 
What will be the saturation current of this diode at 400°K? Assume the rate 
of increase to be constant at the 300°K value. 

PROB. 3.5. The diodes of Prob. 3.4 have 0.2 v forward bias across the 
junction, (a) Find the current at room temperature if the diode is germanium. 

4 Westinghouse R and D Letter, Vol. 4, No. 5, February 1961. 



66 Electronic Engineering 

(b) Find the current at 400°K if the diode is germanium, (c) Find the current at 
room temperature if the diode is silicon, (d) Find the current at 400°K if the 
diode is silicon. Answer: (a) 24.4 mamp (b) 354 mamp (c) 24.4 uamp (d) 345 
mamp. r 

3.5 JUNCTION CAPACITANCE 

A p-n junction and a charged capacitor are similar. As previously 
noted, the stored charge in the region of the junction results from the 
removal of free electrons from the n region, which leaves fixed positive 
donors. Similarly, the filling of the missing covalent bonds of the acceptor 
atoms in the/? material produces fixed negative charges. The removal of 
the free or mobile carriers near the junction produces a depletion region 
which supports fixed excess charges and an electric field (Figs. 3.1 and 
3.7). Some relationships between the barrier potential and the depletion 
width, the junction capacitance and the barrier potential, the maximum 
field intensity and the doping concentrations, and so on will be developed 
with the aid of Fig. 3.7, representing an abrupt junction planar diode of 
unit cross-sectional area. The excess positive charge density in the 
w-doped crystal is approximately equal to qN d and penetrates a distance 
L n (Fig. 3.7b). Similarly, the excess negative charge density in the/?-doped 
crystal is -qN a and it penetrates a distance L v . As indicated by the 
dashed lines in Fig. 3.7, the excess-charge regions do not terminate 
abruptly, but little error will be introduced by assuming abrupt termination 
at the effective distances L n and L p from the junction. Conservation of 
charge requires that the total charge be zero. Then 

qN d L n = qN a L v (3.33) 

The electric field intensity (Fig. 3.7c) may be determined from Gauss' 
law, which states that the total electric flux DA passing through a given 
closed surface is equal to the coulomb charge enclosed by the surface. 
Imagine that a ZY plane is located through the depletion region to the 
left of the junction. The total charge to the left (and to the right) of the 
plane is -qN a (L„ + x). All the electric flux lines originate on positive 
charges to the right of the plane and terminate on the negative charges 
to the left of the plane. Then, since the cross-sectional area is unity, 

S = 5 = _ *N*iL p + x) 

Since the imaginary plane is to the left of the junction, x is negative 
and the electric field reduces to zero at x = -L v . The maximum field 



Diodes 



67 



qN d 



-QNa 



+ + + 
+ + + 
+ + + 
+ + + 



-Ln 



Ln 

I" I I . 

(a) Semiconductor junction 

I 



I 



I 



I 



(b) Electric charge 

! ! ■ 




1 










1 


1 


\ 






-w 






\ 




\ 


> 





Ln 


X *- 




v . J 











(d) Electric potential 

Fig. 3.7. The electric charge, field, and potential relationships in the depletion region 
associated with a p-n junction. 



exists at x — and is given by 



qN a L p qN d L n 



(3.35) 



The electric potential, using the/» material to the left of L„ as the reference, 
is 

V n = -I*"* £dx (3.36) 

J -Lb 

Since the potential difference V h across the barrier is equal to the negative 
of the area under the & curve, inspection of Fig. 3.7c shows that 



1 v h = 



^maxq„ + L„) qN n L v (L v + L„) 



2e 



(3.37) 



68 

But from Eq. 3.33, L n = N a LjN d and 

_ qN a L p \l + NjN d ) 

'h — ' 

Therefore, 



*<,= 



2e 



Electronic Engineering 



(3.38) 



LqN a (l + NjN d )J 
A similar solution for L n yields 



L n = 



2eK 



LqN d (l + NJNJ. 



'A 



l A 



(3.39) 



(3.40) 



Note that the depth of charge penetration, L p or L n , is proportional 
to the square root of the total barrier potential V h and is roughly inversely 
proportional to the square root of the doping concentrations. 

A concentration of charge exists at the junction and also a potential 
difference appears across the junction. Whenever these conditions exist, 
a capacitance also exists. The usual definition of capacitance C is charge 
Q, divided by voltage V. When a change of charge results from a change 
of voltage, the dynamic value of capacitance C may be defined as 



dV 



(3.41) 



For the junction, the charge and voltage are both functions of the distance 
L P or L n . Therefore, Eq. 3.41 may be written as 



C = 



dQjdL p 
dVJdL, 



(3-42) 



A voltage increase d V h will cause an increased depth of penetration dL P 
to the left of L p and an increased depth of penetration dL n to the right 
of L„. But, dQ for the increase to the left of L P is 



or 



dQ = qN a dL t 

7rr qN ° 



We can take the derivative of Eq. 3.38 to obtain 



dK 
dL.„ 



: 2qL P N a 
2e L 



(3.43) 
(3.44) 

(3.45) 



Diodes 



69 



The value of L p from Eq. 3.39 is substituted into Eq. 3.45 to yield 



dVn 



w.( i+! th 



(3.46) 



Now, substitution of Eqs. 3.44 and 3.46 into Eq. 3.42 yields 



C = 



qN a 



2qN a (l + NjN d )V h 



qN a e 



.2(1 + NJN d )V h . 



'A 



J 

Equation 3.47 can be written as 

C ~ KVJ, 
where K is a constant. 



farad/m 2 (3.47) 



(3.48) 



PROB. 3.6. A given silicon diode has N a = 10 16 atoms/cm 3 and N a = 10 17 
atoms/cm 3 . Determine the effective length of the depletion region on each 
side of the planar junction in a silicon diode with 20 volts reverse bias 
applied. Assume the zero bias barrier voltage V h0 ~ 0.65 v at this doping concen- 
tration, e = 1.06 x 10^ 10 f/m for silicon. Answer: L p =1.58 x 10^ cm, 
L n = 1.58 x 10-* cm. 

PROB. 3.7. What is the junction capacitance of the diode of Prob. 3.6 if the 
cross-sectional area of the junction is 10~ 6 m 2 ? 



3.6 JUNCTION BREAKDOWN 

The breakdown effect shown at point c in Fig. 3.5 will now be considered. 
One theory attributes the rather abrupt current increase to the high 
potential gradient (or electric field) which exists at the junction. According 
to this theory, the high electric field is able to disrupt the covalent bonds 
and therefore greatly increase the minority carriers. This effect which was 
proposed by Zener is known as Zener breakdown. Another accepted 
theory for the voltage breakdown is the avalanche breakdown. This theory 
was founded by Townsend while he was studying the behavior of gases 
subject to electron bombardment. Accordingly, this theory was originally 
applied to gaseous conduction but applies just as well to the semiconductor. 

According to the avalanche theory, a few carriers are generated in the 
intrinsic semiconductor material owing to thermal action, as previously 
discussed. These carriers are accelerated by the high electric field near the 
junction until high velocities are acquired. A carrier with sufficient energy 
can produce an electron-hole pair when this carrier collides with a neutral 
atom. The new carriers so produced are free to be accelerated and in 
turn to produce additional carriers. The origin of the term avalanche can 
be seen by the foregoing explanation. 



70 Electronic Engineering 

Since the electrons are much more mobile than the holes, most of the 
carriers are produced by electron collisions. The electrons have many 
random collisions as they travel through the semiconductor. In order for 
the avalanche effect to manifest itself, the electrons must obtain sufficient 
energy in traveling one mean free path to produce ionization of the atoms 
in the semiconductor. Hence, the electrons must have a kinetic energy 
equal to or greater than the gap energy of the semiconductor for an 
avalanche to be produced. 




^Breakdown of ' Breakdown of 
a typical diode ,^a more heavily 
doped diode 



Fig. 3.8. The effect of impurities on the breakdown potential of a p-n junction. 

The breakdown potential is a function of the impurity concentrations 
of the semiconductor. As can be noted by Eqs. 3.39 and 3.40, the depletion 
width varies inversely with the impurity concentration. Hence, as the 
impurity concentration increases, the depletion region becomes thinner. 
As the depletion region becomes thinner, the electric field intensity becomes 
higher for a given junction voltage. Higher electric field potentials produce 
higher electron energies per mean free path. Therefore, a heavily doped 
p-n junction will have a lower breakdown potential than a relatively lightly 
doped p-n junction. This effect is illustrated in Fig. 3.8. 

Many diodes are designed and constructed for operation on the ava- 
lanche portion of the characteristic curve. These diodes are known as 
zener diodes or reference diodes. These diodes operate in a region (from 
c to don the curve of Fig. 3.8) where the current is essentially independent 
of voltage. The uses of these diodes are discussed in Section 3.15. 

Reference diodes are available with breakdown voltages ranging from 
about 3 v to well over 100 v. The avalanche appears to be the primary 
breakdown mechanism in diodes with reference voltages above about 7 v 
because the breakdown voltages of these diodes increase with temperature. 



Diodes 71 

The reduced mean free path at elevated temperatures would account for 
this positive temperature coefficient. In contrast, diodes with breakdown 
voltages less than about 6 v have a negative temperature coefficient, which 
indicates that the zener breakdown mechanism is predominant in this 
range. The increased kinetic energy of the valence electrons would aid the 
high field in producing carriers and thus cause a negative temperature 
coefficient. Diodes which break down at about 6 to 7 v have essentially 
zero-temperature coefficient. 

PROB. 3.8. Zener breakdown will occur in a silicon diode if the electric field 
intensity exceeds about 10 8 v/m. A given silicon diode has N a = N d = 10 18 
atoms/cm 3 . Will zener breakdown occur with 5.0 reverse bias applied to this 
diode? K Ao ~0.9v, e = 1.06 x 10- 10 f/m. At what voltage will a zener 
breakdown occur? 

3.7 THE DIODE SYMBOL 

The symbol of a p-n junction diode is shown in Fig. 3.96. The polarity 
signs are shown for forward bias. Notice the arrowhead points in the 
direction of conventional current (as opposed to electron flow) when the 
diode is forward biased. '- 



(a) Actual diode 



v-e 



(b) Symbol 
Fig. 3.9. The p-n diode symbol. 



3.8 THE TUNNEL DIODE 

The tunnel diode is a junction diode which is very heavily doped with 
impurities. In fact, a good tunnel diode must contain impurity concentra- 
tions greater than 5 x 10 19 /cm 8 for silicon and greater than 2 x 10 19 /cm 3 
for germanium. 5 This heavy concentration of impurities affects the elec- 
trical characteristics of the junction. As noted in Section 3.6, heavy doping 
reduces the thickness of the depletion region. Actually, the depletion 
width 8 will be in the order of 100 A (1 A = lO" 8 cm) when the doping 

6 1. A. Lesk, N. Holonyak, and U. S. Davidsohn, "The Tunnel Diode-Circuits and 
Applications," Electronics, November 27, 1959, Vol. 32, No. 48, pp. 60-64. 

•Bernard Sklar, "The Tunnel Diode— Its Action and Properties," Electronics, 
November 6, 1959, Vol. 32, No. 45, pp. 54-57. 



72 



Electronic Engineering 



concentration is in the order of 10 20 atoms/cm 3 . To enhance this thin 
depletion region, the transition from n to p material must be as abrupt 
as possible. 

As was previously mentioned, the heavy doping reduces the junction 
breakdown voltage. When the concentration of impurities reaches the 
values listed previously, the semiconductor begins to resemble an alloy 




Fig. 3.10. The characteristic curve of a typical tunnel diode. 

and the breakdown voltage is reduced essentially to zero. 

An additional effect is to place the Fermi level of the n material in the 
conduction band. In addition, the Fermi level of the p material is in or 
is very near the valence band. 

As a result of the foregoing effects, the voltage-current characteristics 
of the tunnel diode are as shown in Fig. 3.10. The reason for this unusual 
shape will now be considered. 

As noted before, the Fermi levels in the « and/? material must coincide 
in a diode with no external voltage. Accordingly, the energy diagram 
for no external voltage would be as shown in Fig. 3.11a. According to 
the "classical" concepts, the only electrons able to cross the junction are 
the few electrons with sufficient energy to climb the potential barrier. 
(This potential barrier is higher in heavily doped tunnel diodes than in 
conventional p-n junctions because of the location of the Fermi level.) 
However, according to quantum mechanics there is a probability that 
certain electrons which do not have enough energy to climb the potential 



Diodes 73 

hill can pass from the p to the n material (and also from the n to the p 
material). These electrons pass between the valence band of the/) material 
and the conduction band of the n material. The electrons are said to 
tunnel through the potential barrier and hence the name tunnel diode. 
The quantum mechanics, therefore, predicts that current flow can occur 
in Fig. 3.1 la. However, the probability that electrons will cross from the 
p to the n region is the same as the probability that electrons will cross 
from the n to the p region. Consequently, the two currents cancel in 
Fig. 3.1 la. This condition corresponds to the origin (point A) of the curve 
of Fig. 3.10. 

When a small forward bias is applied across the diode, the energy 
relation is given by Fig. 3.1 lb. Here there is a greater probability that the 
electrons will pass from the n to the p region than from the p to the n 
region. Consequently, a net flow of current occurs across the junction. 
This condition is shown as point B on the curve in Fig. 3.10. (This 
current is known as Esaki current after the Japanese inventor of the tunnel 
diode.) 7 

If the forward bias is increased beyond the level just considered, some 
of the electrons in the conduction band of the n material have the same 
energy as the forbidden band in the p material. These electrons cannot 
exist in the forbidden band and, therefore, must stay in the n material. 
Consequently, the number of electrons available for transfer across the 
junction is reduced. This action accounts for the region B to C in the 
characteristic curve of Fig. 3.10. In this region, an increase of bias causes 
fewer electrons to be available for transfer across the junction. Conse- 
quently, more forward bias results in less current across the junction. 
Finally, the condition is reached where most of the conduction-band 
electrons in the n material have the same energy as the forbidden band in 
the/? material. This condition is shown in Fig. 3.11c, and the corresponding 
current is indicated by point C in Fig. 3.10. 

As the bias is increased beyond point C (Fig. 3.10), more of the electrons 
in the n material are able to "climb" the potential barrier so the action is 
the same as in conventional diodes. This condition is shown in Fig. 3.1 la" 
and as point D on the current curve of Fig. 3.10. 

Actually, the most useful section of the characteristic curve is the region 
from B to C, where an increase of voltage causes a decrease of current and 
the device behaves as if it were a negative resistance. The usefulness of this 
negative resistance characteristic is indicated in Probs. 3.11 and 3.12. 

' The actual theory is much more involved than this description indicates. For 
example, the momentum of a tunneling electron must be the same on each side of the 
forbidden band. Momentum differences cause the diode material to vibrate mechani- 
cally. 



74 



Electronic Engineering 



p-type 



n-type 



S3 



Conduction 
band 


3 




Forbidden 
band 


i 


Conduction 
band 


Empty states 

Valence 
band 


-V- 




\ \ > 


Electrons 


\ Vi 




\ 

1 


Forbidden 
band 


Valence band 



Distance - 
(a) 



Fermi 
level 



p-type 



n-type 



c 

Conduction 7 _ . . 
band Conduction 
| band 


Forbidden V 
band V_ 


\ Electrons 


Empty states \ 


\ band 


Valence 

i Valence 
| band 

i 



Distance »- 

(c) 

Fig. 3.11. Energy relationships in a tunnel diode, (a) No external voltage; (b) maxi- 
mum Esaki current; (c) zero Esaki current; (rf) regular junction current. 



Diodes 



75 





P-type •§ ra-type 

c 

3 

Conduction "? 
band 
Conduction 


u 


\ band 

Forbidden \, 
band \ 






KB 




01 

c 


\ 
\ Forbidden 

., . I\ band 
Valence \ 

band | \ 




Valence 
band 



Distance — 
(b) 



p-type 



n-type 



Conduction 
band 


c 

3 

7\ 


Conduction 
band 


Forbidden 
band 


\ v 




Electrons 






Empty states 


^ 


Forbidden 


Valence 
band 


\ 




i 


Valence 
band 



Distance 
id) 



76 Electronic Engineering 

In addition to the negative resistance characteristic, the tunnel diode has 
other advantages. One of the most important advantages is the speed with 
which electrons cross the junction. In the conventional junction diode the 
electrons diffuse through the junction at a velocity which is considerably 
below the velocity of light. Consequently, the time required to cross the 
depletion region is relatively great. In contrast, the electrons in the tunnel 
diode exhibit their wave characteristic and cross the very thin depletion 
region with a velocity approximating that of light. 8 As a result of this 
very short transit time, the tunnel diodes can be used at very high fre- 
quencies. In fact, the tunnel diodes have been used in circuits with fre- 
quencies 9 above 10 GHz. Other advantages of the tunnel diode include 
small size, ruggedness, and ability to operate in high radiation fields and 
at high temperatures. 

The tunnel diode also has disadvantages. One obvious disadvantage (Fig. 
3.10) is the low-voltage region over which the device operates. (This low 
voltage can be an advantage as well as a disadvantage, however.) The main 
disadvantage is the lack of isolation between input and output circuits. The 
advantage of this isolation will become evident as the chapters on amplifiers 
are studied. 

PROB. 3.9. What is the conductivity of n germanium at room temperature if 

2 x lO 19 impurity atoms are present in each cubic centimeter? Compare with 

Prob. 2.6. Answer: 1.25 x 10 6 mhos/m. 

PROB. 3.10. Determine the width of the depletion region in a silicon diode 

which has both n and/> regions doped to a concentration of 5 x 10 20 impurity 

atoms per cc. Assume V h0 = 1.14 v. 

PROB. 3.11. A battery with a voltage V B is connected in series with a 5-ii 

resistor, a 5-mh inductance, and 0.05-/<f capacitance. 

(a) Determine the voltage across the inductance as a function of time if the 
circuit is connected together at time t = 0. 

(b) Add a negative resistance of 5 CI in series with the circuit in part a and 
then find the voltage across the inductance as a function of time. 

(c) What use does this problem suggest for a tunnel diode? 

PROB. 3.12. An a-c voltage generator has an RMS open terminal voltage of 
0.01 v and an internal resistance of 10 CI. The generator is connected to a 
resistive load of 100 CI. 

(a) Find the voltage across the load, the current through the load, and the 
power delivered to the load. 

8 This statement may seem to be in conflict with the principles stated in Chapter 1 . 
However, an electromagnetic wave is able to move along a conductor at approximately 
the speed of light even though the electrons in the conductor move at a relatively slow 
speed. The effect is transferred down the conductor from carrier to carrier much faster 
than the carriers actually move. The same principle applies to the tunneling effect. 

• The abbreviation GHz means gigahertz where giga (abbreviated G) is 10» and 
Hertz (abbreviated Hz) is cycles per second. This is standard IEEE notation and will be 
used in the remainder of the book. 



Diodes 



77 



(b) Add a negative resistance of 105 ft in series with the circuit in part a. 
Find the voltage across the load, the current through the load, and the power 
delivered to the load under these conditions. 

(c) What use does this problem suggest for a tunnel diode? 



3.9 HIGH-VACUUM DIODES 



Plate 



Insulated 
filament 



Cathode 



The high- vacuum diode is constructed as shown in Fig. 3.12. The 
important components of this vacuum diode are : a heater wire, an electron- 
emitter surface, and an element known as a plate. This entire structure 
is placed in a glass or metal envelope. Most 
of the air in the envelope is removed so the 
active elements are surrounded by an effec- 
tive vacuum. It is impossible to remove all 
of the air in the envelope, but the pressure 
inside the envelope is reduced to the vicinity 
of 10~ 6 mm Hg (Atmospheric pressure is 760 
mmHg.) 

The heater wire is a piece of resistance 
wire and is heated by an electric current 
which flows through the wire. In some 
vacuum tubes, the heater wire and the elec- 
tron-emitter surface are combined. In this 
case, the hot heater wire or filament emits the 
electrons. This type of emitter is known as a 
filament-type emitter. Other tubes use the 
heater wire or filament to heat a metal sleeve 
(Fig. 3.12). The metal sleeve, which is usually 

coated with a good thermal electron emitter such as the rare-earth oxides 
is known as a cathode. The cathode is usually electrically insulated from 
the heater. 

The heated cathode emits electrons into the vacuum. If the plate is 
more po sitive than th>» cat hode, these electrons are attracted to the plate. 
As ac onsequence, current flows through the tub e. If the jriate is more 
negative than the cathode, the electrons are repelled back to the cathode 
a nd no current flows through the tube. The action is similar to the p-n 
junction action except the only carriers are electrons. A second difference 
exists because the plate cannot emit electrons at normal temperatures so 
no reverse current flows in the vacuum tube. In fact, since the electrons in 
actual tubes are emitted from the cathode with a finite velocity, some 
current flows in the forward direction for zero and even slightly negative 
plate voltages. 




Fig. 3.12. The high-vacuum 
diode. 



78 Electronic Engineering 

3.10 VACUUM DIODE CHARACTERISTIC CURVES 

A plot of plate current as a function of plate voltage for a high- vacuum 
diode is given in Fig. 3.13. When the plate voltage is more than one or two 
volts negative, the current is zero. As the plate is made more positive, 
the current increases exponentially until point A is reached. At point A, 
saturation of the plate current occurs. This saturation exists because the 



* 




Fig. 3.13. Plate current vs plate voltage of a vacuum diode. 

plate is collecting all the electrons the cathode is able to emit. If the 
temperature of the cathode is increased, the plate current increases to a 
higher value before saturation occurs. This behavior is shown by the T 2 
curve of Fig. 3.13, where saturation exists at point B on the curve. 

The value of saturation current for a given temperature can be found 
from a relationship known as Richardson's equation. 



J = A T 2 e- w »' kT 



(3.49) 



where / = the temperature-limited current density of the cathode 

A a = an experimentally determined constant (different for each 

metal) 
T = the temperature of the cathode in °K 
k = Boltzmann's Constant 
W w = the work function of the cathode surface. (This W h is the 
amount of energy an electron must have in order to escape 
from the cathode.) 
This equation is very similar to the equation which gives the number of 



Diodes 79 

conduction-band electrons in a semiconductor material (Eq. 2.28). Since 
Eq. 2.28 gives the number of electrons in a material with sufficient energy 
to become carriers at a given temperature, and since Eq. 3.49 determines 
the number of electrons in a cathode with sufficient energy to escape from 
the surface of the cathode at a given temperature, the equations should be 
similar. 
The constants for a few common materials are given in Table 3.2. 

TABLE 3.2 
Constants for a Few Common Materials 

A W w Melting 

Temperature 
Material [amp/m 2 -(°K) 2 ] [ev] [°K] 



Cesium 


16.2 x 10 4 


1.81 


299 


Copper 


65 x 10 4 


4.10 


1356 


Nickel 


30 x 10 4 


4.6 


1725 


Thorium 


60 x 10 4 


3.40 


2118 


Tungsten 


60 x 10 4 


4.52 


3643 


Thorium on tungsten 


3.0 x 10 4 


2.60 




Oxide coated 


0.1 x 10 4 


1.00 





PROB. 3.13. Calculate the saturation current for an oxide-coated cathode with 
an area of 2 cm 2 and a temperature of 300°K (room temperature). Repeat for 
a temperature of 1 100°K. 

PROB. 3.14. Calculate the saturation current for a 2 cm 2 tungsten emitter at 
2500°K. Answer: 0.571 amp. 

The actual plate current vs plate voltage curve (Fig. 3.13) does not 
exhibit true saturation. This actual curve digresses from the theoretical 
curve as a result of the Schottky effect . T his Schottky effect is produced by 
th e potential gradient at the cathode^ This potential gradient becomes 
positive at the cathode when saturation is reached and increases as the 
plate potential increases. This positive gradient at the cathode helps 
s ome elec trons, which have kinetic en ergies less than the work function 
of the cathode, t o overcome the surface barrier of the cathode. Hence, 
as the plate becomes more positive , more electrons are "helped oii ^of 
the cathode ana emission increases. 

Commercial tubes are designed so that the saturation current is much 
higher than the maximum recommended plate current. Hence, if com- 
mercial tubes are operated at the manufacturers' suggested filament 
voltage, the tube will not be operated in the saturation region. 

As Fig. 3.13 indicates, the current below saturation does not increase 



80 



Electronic Engineering 



linearly with voltage but follows an exponential curve. The mutual 
repulsion of the electrons in the space between the cathode and plate is 
responsible for the exponential shape of the current vs voltage curve. A 
simple derivation will illustrate this behavior. 

Assume that two infinite parallel plates exist. One plate represents the 
cathode, and the other plate represents the plate. A section of this 



Cathode 




+ Volts 



Fig. 3.14. Field distribution of a diode with a charge between the plates. 



configuration is shown in Fig. 3.14. Also shown in Fig. 3.14 is a small 
volume of space between the plates which contains a charge density p. 
From Appendix I, 

P 



dx dy dz e 



(1.8) 



For the field configuration shown in Fig. 3.14, -^ = —^ = 0. Therefore, 
Eq. 1.8 becomes dz dy 



di x -d 2 V p 



(3.50) 



dx dx 2 e v 

The electric field $ is equal to —dVjdx and the value of e for a vacuum 



is e„. 



The second equation required is given in Chapter 1 as Eq. 1.12 



kmv' 



Vq 



(1.12) 



where V is the voltage through which the electron has been accelerated. 



Diodes 81 

In this case, the charge on an electron q has been inserted in place of the 
general charge Q of the general Eq. 1.12. 
The third relationship is 

" = -P (3.51) 

v 

where J = the current density in amp/m 2 

v = the velocity of the electrons which constitute the current in 
m/sec 
—p = the electronic charge density in coulombs/m s 
When the value of v as given by Eq. 1.12 is substituted into Eq. 3.51, 

-P = ^4 (3.52) 

{2Vq) A V 

The quantity p is negative for the space current of electrons. The value of 
p as given by Eq. 3.52 can be substituted back into Eq. 3.50 to yield 



dx 2 e v \2q 
This equation is difficult to integrate. However, 



= L(m.f v -H (353) 



\dx! 



= 2 dV<?L ( 3.54) 

dx dx dx 

Now, if both sides of Eq. 3.53 are multiplied by 2(dV[dx), Eq. 3.53 becomes 

2 ^ = 2A(m.y v -*dv 

dx dx 2 € \2ql dx V ' 

Both sides of this equation can now be integrated to give 

At the cathode, V is zero and § is assumed to be zero due to the space 
charge. 10 Since £ is equal to —dVjdx, K x must be equal to zero. Taking 
the square root of both sides of Eq. 3.56, we find that 



dV = ["4j/m\ V 
dx le v \2q! 



V Vi (3.57) 



10 This development assumes that the electrons leave the cathode with zero velocity. 
Actually, the electrons leave the cathode with a wide distribution of velocities. Hence, S 
is not zero but actually a small negative value. The effect of ignoring the initial velocity 
of the electrons can be seen by referring to Fig. 3.15. 



82 

or 



v-« iy - \±[f ff dx 



■„ \2q) 
Integration of both sides of Eq. 3.58 yields 



Electronic Engineering 



(3.58) 



iv v < = 



"4//mV 



}A-\ 'A 



x+ K 2 



(3.59) 



But, from Fig. 3.14, x is zero at the cathode. Since the voltage at the 
cathode is also zero, AT 2 must be zero. 




Fig. 3.15. A plot ofFvsa; as given by Eq. 3.62 and as exists in an actual diode. 

A plot of V as a function of x (as given by Eq. 3.59) is shown as the 
theoretical curve in Fig. 3.15. This plot is seen to be different from the 
curve in an actual diode. The discrepancy between the two curves is due 
to the initial velocities of the electrons from the cathode. These electrons 
form a "cloud" of electrons, which is known as a space charge, a short 
distance from the actual cathode. The actual current in the tube is drawn 
from this electron cloud. Hence, a "virtual cathode" appears to exist a 
short distance from the actual cathode. This virtual cathode is at a more 
negative potential than the actual cathode. However, the curve from the 
virtual cathode to the plate does have the form given by Eq. 3.59 if the 
potential of the virtual cathode is used as the reference (V = 0) point. 



Diodes 83 

Substituting zero for K 2 in Eq. 3.59 and solving for /, we find that the 
current density is 

4e V* 
■/= * ,^ -V arnp/m 2 (3.60) 

9{ml2q) A x 2 

When the values of e„, m, and q are substituted into Eq. 3.60, the 
current density becomes 

J = amp/nr (3.61) 

x 2 

where x is the distance from the cathode to the point in question in meters 
and V is the potential in volts x meters from the cathode. For a diode with 
a distance d meters between the cathode and the plate, the current density 
at the plate is 

= 2.335 x 10-F? 
d 2 

where V p is the potential difference between the plate and the cathode in 
volts. 

This relationship is known as Child's Law. Equation 3.60 states that 
the current density is proportional to the voltage to the f power at any 
given distance from the cathode. The fact that the amount of current 
leaving the cathode must be equal to the amount of current reaching the 
plate for steady-state conditions indicates that the current does not vary 
with x. Therefore, the potential is seen to vary as the f power of x. 
(If Eq. 3.61 is solved for V, x appears to the f power.) This derivation 
was for a plane electrode configuration, and a similar derivation for 
cylindrical electrodes leads to a similar but more complicated relation- 
ship. 11 Even in the cylindrical configuration, however, the current is 
theoretically proportional to voltage to the f power. Hence, the expo- 
nential shape of the current vs voltage curve (Fig. 3. 13) is due to the charge 
on the electrons which are in transit from the cathode to the plate. 

PROB. 3.15. Justify Eq. 3.51. 

PROB. 3.16. A plane electrode diode has the plate 5 cm from the cathode. A 
voltage of +100 v is applied between the plate and the cathode, (a) What is the 
current density at the plate? (b) If the tube is 1.5 cm by 5 cm, what is the total 
current of this tube ? (c) If the plate voltage is raised to 200 v, what will be the 
plate current? Answer: (a) 0.93 amp/m 2 (b) 0.7 mamp (c) 1.98 mamp. 
PROB. 3.17. A plane electrode diode has 1 cm distance between the plate and 
the cathode. The plate and cathode are each 1 cm by 5 cm. Plot the curve of 
plate current vs plate voltage for this diode from to 200 plate v. (Use 
steps of 50 v.) 

11 K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, New York, 1948, pp. 173-175. 



84 Electronic Engineering 

PROB. 3.18. Plot the theoretical curve of voltage vs distance from the cathode 
for Prob. 3.17 if the plate is +100 v and the cathode is v. 

3.11 SYMBOLS FOR VACUUM DIODES 

The symbols for the two types of vacuum diodes shown in Fig. 3.16 
represent the actual element arrangements inside the tubes. Tube manuals, 
which can be purchased at most radio supply stores, list socket connections, 
characteristic curves, voltage and current ratings, and so forth, for most 
commercial tubes including diodes. 




Fig. 3.16. Symbols for vacuum diodes (a) filament-type emitter (6) cathode-type 
emitter. 



3.12 GAS DIODES 

A gas diode may be constructed in the same manner as a vacuum diode. 
However, instead of maintaining a vacuum inside the envelope, the gas 
diodes have a carefully controlled amount of a pure gas present. 

The exact behavior of a gaseous medium as an electrical conductor is 
very complicated. 12 In fact, the conductivity of the gaseous medium is a 
function of the gas pressure, the gas temperature, the material from which 
the electrodes are made, the magnitude of the applied voltage, and even 
the nature of the voltage source. As a result of this complication, the 
following discussion of the gaseous conduction process is somewhat 
superficial and oversimplified. 

When the plate is more than a volt or two negative with respect to 
the cathode, no current flows in the gas diode. The electrons are then 
repelled back to the cathode as in the high-vacuum tube. When the 
plate voltage is made positive, electrons from the cathode are attracted 

12 A comprehensive treatment of gaseous conduction is given in Gaseous Conductors 
by James D. Cobine, Dover Publications, New York, 1958. 



Diodes 



85 



to the plate. These electrons must pass through the gas molecules on the 
way to the plate. Although the concentration of gas atoms is not nearly 
so great as is the concentration of atoms in a semiconductor, the action 
is somewhat the same. 

If electrons are emitted from a heated cathode, these electrons diffuse 
through the gas molecules. This diffusion process tends to create a uni- 
form electron distribution throughout the volume. If a plate of zero 







/ 






/ 


4 




/ 


| 




/ 


i 




/ 






/ 


E 
.2 




/ 's^ High vacuum 




/ characteristic 






c / 


E 


v / 


-2 


\ / 


IS 


1 / 


Q- 


// 




// 


A 


*Sb 



Plate volts ■ 



Fig. 3.17. A plot of current vs voltage for a hot cathode gas diode tube. 

potential is inserted in this gas, each electron that strikes the plate will 
be captured. Consequently, a very weak current will flow to a plate of 
zero potential from a heated cathode. This condition is shown as point 
A of Fig. 3.17. 

As the plate is made more positive, electrons drift through the gas 
molecules to the plate. These electrons will have many collisions with the 
gas molecules. However, as long as the energy of the electrons is small, 
each of these collisions will be elastic. (In an elastic collision, the electron 
has essentially the same kinetic energy after the collision as before the 
collision.) The current increases exponentially with voltage as shown from 
A to B in Fig. 3.17. 

At point B of Fig. 3.17, the current digresses from the simple exponential 
curve of A to B. At this point, some of the electrons have enough energy 
to ionize the gas molecules on collision. The gas molecule receives enough 
energy from the free electron to allow a valence electron (in some cases 
more than one valence electron) to escape. As a result, the gas molecule 
becomes a positive ion. The electrons continue on to the plate and the 
positive ion travels toward the cathode. Because of its comparatively 
large mass, the positive ion travels much more slowly than the electron 
does. Therefore, the ionized gas acquires a positive charge which assists 
in the acceleration of the electron away from the cathode. 



8* Electronic Engineering 

As the current density is increased, the voltage rises slightly at first, but 
then as additional current is drawn more ionizing collisions occur and the 
voltage across the diode decreases until the voltage is essentially equal to 
the ionization potential (usually about 12 v) of the gas. This condition is 
indicated as point C in Fig. 3.17. An increase of current beyond point C 
causes essentially no increase of voltage across the tube. (Actually the 



{ 



z* c 



E 



l_p / 



Plate voltage - 



Fig. 3.18. The characteristic curve of a cold cathode gas tube. 

voltage will increase if the current is raised to a high enough value, but 
this high value of current is beyond the maximum current rating of the 
diode.) 

A second type of gas diode, also in wide usage, contains a plate and a 
cathode surface, but the cathode is not heated. Consequently, the tube is 
referred to as a cold cathode diode or (because of its principal usage) as a 
voltage regulator diode. The voltage-current characteristics of this type of 
tube are shown in Fig. 3.18. 

In order to produce ions in the cold cathode tubes, a small amount of 
radioactive material is included inside the tube. 13 The radiation from this 
material produces positive and negative ions in the tube. Consequently, 
a small voltage across the tube attracts positive ions to the negative 
cathode and negative ions (or electrons) to the positive plate. When all 
of the ions produced in the tube are collected, the current through the tube 
reaches a saturation condition as indicated at point A in Fig. 3.18. At 
point B in Fig. 3.18, the current starts to increase with increasing potential. 

13 Glenn Cassidy, "Radioactive Krypton in Cold-Cathode Gas Tubes," The Western 
Electric Engineer, Vol. IV, No. 1, January 1960. 



Diodes 87 

At this point, some of the electrons have enough energy to ionize the gas 
molecules on collision. The gas molecule is broken down to an electron 
(in some cases more than one electron) and a positive ion. The electrons 
continue on to the plate and the positive ion travels back to the cathode. 
As a result of this ionizing process, the current increases exponentially. 
This ionizing (voltage) region was first investigated by J. S. Townsend and 
accordingly is known as the Townsend I Region. The Townsend I region 
is shown in Fig. 3.18 as the region between B and C. 

In gaseous conduction, the negative carriers are free electrons as in the 
semiconductor. In contrast, the positive carriers are the positive ions 
instead of the holes. However, because of the much greater mass of the 
positive ions (as previously mentioned), the positive carriers move at a 
much slower velocity than the electrons do. 

Another phenomena occurs at point C of Fig. 3.18. At this point, the 
actual current again digresses from the expected curve. In this region the 
positive ions have enough energy when they strike the cathode to cause 
secondary electrons to be emitted. 14 As a result, the current increases 
above the value expected for pure gaseous conduction. This region is 
known as the Townsend II region and is shown in Fig. 3.18 as the region 
between C and D. 

The phenomenon from the origin to D of Fig. 3.18 is known as a dark 
discharge; that is, there is no visible light emission from the tube. If the 
radiation is removed, the current drops back to zero. However, at point D 
a new phenomenon occurs. The ions produce a visible glow in the gas and 
the discharge becomes self-sustaining. In this case the number of secondary 
electrons from the cathode is sufficient to keep the current flowing. Hence, 
current is maintained even if the radiation is removed. 

The self-sustaining condition is known as a glow discharge. The total 
voltage actually decreases across the tube as shown in the curve (part D to 
is) of Fig. 3.18. From point Zi to F of Fig. 3.18, the voltage remains almost 
constant whereas the current changes over quite a wide range. The change 
of current from E to F is much, much greater than the change of current 
from the origin to D. The current at point E flows through a restricted 
volume of the gas. As the current increases, the active volume of gas 
increases until point F is reached. When the tube is operating at point F, 
the entire volume between the cathode and the plate is active. When the 
current is increased beyond point F, the voltage across the tube increases 
again. However, if the voltage is increased very much beyond point F, an 
arc discharge occurs (much as in electric arc welding) and the tube is 
quickly destroyed because of excessive heat. 

14 Positive ions usually do not achieve sufficient speed to cause significant ionization 
in this voltage range. 



88 



Electronic Engineering 



Most cold cathode gas diodes are designed to be operated in the E 
to F region (Fig. 3.18) of the characteristic curve. In this region, the 
current is almost independent of voltage. Typical tubes have operating 
potentials from 75 v for some tubes to 150 v for other tubes. The type of 
gas in the tube and the pressure of the gas determines the operating 
potential. The potential at point D of Fig. 3.18 is known as the striking 
potential. The voltage across the tube must rise to this striking potential 
before the tube can drop back to the operating potential. 




(a) Heated cathode 
tube 



(6) Cold cathode 
tube 



Fig. 3.19. Schematic representation of a gas diode. 

The symbolic representations for gas diode tubes are shown in Fig. 3.19. 
The symbol for the heated cathode type is the same as the symbol for a 
high-vacuum diode except a small black dot is included in the envelope to 
represent the gas. 



3.13 GRAPHICAL SOLUTION OF DIODE CIRCUITS 

In order to use diodes intelligently, methods of analysis and design must 
be understood. In general, two basic approaches will be described. One 
approach (described in Section 3.14 and Section 3.15) uses an equivalent 
circuit for the diode. However, in this section the diodes and associated 
circuit elements will be analyzed by graphical methods. 

In the graphical method of solution, the current vs voltage charac- 
teristics of the diode are expressed in graphical form. Frequently the tube 
manuals and manufacturers' data sheets present the characteristics of the 
diodes in graphical form. These plots of current vs voltage are known as 
characteristic curves. The characteristic curves furnished by the manu- 
facturer represent average diodes of a given type. Hence, if very accurate 
curves are required, the characteristic curve for the particular diode in 
question must be plotted from measured voltages and currents. 



Diodes 



89 



With the diode's characteristics presented in graphical form, the current 
vs voltage characteristics of the circuit external to the diode must be 
presented in graphical form. Intersections of the various curves represent 
solutions of the circuit currents and voltages. An example will illustrate 
this procedure. 

Example 3.1. A circuit is connected as shown in Fig. 3.20. Find the output 
voltage v if vj has the form shown. The characteristics of a 5U4 tube are given 
in Fig. 3.21. 




\ of 5U4 ? y 



— i 

i i 

— i — i — i 

10 20 30 40 50 60 70 80 90 100 
Time in milliseconds 



500 a' 



Fig. 3.20. A typical diode circuit. 

When vj is v, essentially no current flows through the circuit. Hence, v Q is 
if V[ is 0. However, when v f is +50 v, some current will flow through the cir- 
cuit. From Fig. 3.20, the voltage equation around the plate circuit loop is 



v P = v f — i P R L 



(3.63) 



where v t = the instantaneous input voltage 

v P = the instantaneous voltage across the diode (plate voltage) 
i P = the instantaneous current through the loop (plate current) 
This equation contains two unknowns i P and v P . However, a second relation- 
ship between i P and v P is given by the diode characteristic curve (Fig. 3.21). 
Hence, enough information is available to solve for both i r and v P . 

Since the characteristic curve is a graphical representation of the relationship 
between i p and v P , it would be desirable to find a graphical representation of 
Eq. 3.63. Fortunately, if v p and i P are the variable quantities of Eq. 3.63, this 
equation represents (graphically) a straight line. This line will have a slope of 
— 1 //Randan intercept on the o P axis (point A) of v t . Accordingly, iff/ is 50 v, 
the line ABC represents Eq. 3.63. In this case, the intercept with the "v P axis 
(point A) is 50 v. A second point on this line is found by noting that if v P = 0, 



ip 



Hi. 



(3.64) 



Here, with v l of 50 v and R L of 500 Q, i P is 100 ma and is indicated as point C 
in Fig. 3.21. 



90 



Electronic Engineering 



250 



200 



!150 



fclOO 



50 



F 










\ 
X 
X 
X 
X 


\ / 








C 


^1e 
/ X 


\ 

X 






X 
\ 
X 
X 


(b 


\ 

\ 


\ 
\ 
\ 






X 
X 
\ 




X 


X 
N x* 



20 40 60 80 

Plate voltage in volts 



100 



Fig. 3.21. The characteristic curve of a 5U4 tube. 



100 
50 

V) 

o 

E ° 

-50 
-100 



=2 70 h 

O 

I 30 




I 



I 



H — h — — ! — 4— L 

1 ! 

! ! ! 






— i — , 



i 
i 
i 
i i 



10 20 30 40 50 60 70 80 90 100 110 120 
Time in milliseconds 



Fig. 3.22. The voltages v, and v for the circuit of Fig. 3.20. 



Diodes 91 

There is only one point (point B) common to both the characteristic curve 
and line AC. Hence, point B is the only solution for Eq. 3.63 that also satisfies 
the voltage-current requirements of the diode. Thus, i P is 60 ma and v P is 20 v 
when Vj is +50 v. In this case, v is equal to i P times R L or v = 30 v. 

Similarly, line DEF represents Eq. 3.63 when Vj is 100 v. Point E is the 
required operating point with i P = 140 ma and v of 500 x 0.14 or 70 v. 

Whenever the plate of the diode is more negative than the cathode, essentially 
no electrons will pass through the tube. Hence, v = for all negative values of 
plate voltage. Accordingly, the relation between v 2 and v Q is as shown in Fig. 3.22. 

The foregoing circuit is known as a clipper circuit, since the negative 
portion of the input voltage waveform was clipped off. If the plate and 
cathode of the diode are interchanged in the circuit of Fig. 3.20, the 
positive portion of the input wave will be clipped. 

PROB. 3.19. A circuit is connected as shown in Fig. 3.20. If R L is changed to 
1000 n, plot v . Repeat if R L is increased to 10,000 fl. 

PROB, 3.20. A circuit is connected as shown in Fig. 3.20. What value must 
R L have if vo is 80 v when vi is 100 v? 

3.14 DIODE EQUIVALENT CIRCUITS 

As mentioned in Section 3.13, an equivalent circuit can be used for anal- 
ysis and design of diode circuits. The diode is replaced by linear circuit 
elements (resistors, capacitors, batteries, etc.) and switches the combination 
of which has approximately the same current vs voltage characteristics as 
the diode. With the equivalent circuit used in place of the diode, standard 
circuit analysis can be used to determine the circuit performance. 

To illustrate the process of obtaining a suitable electrical network, 
consider the characteristic curve of the 6AL5 vacuum diode shown in 
Fig. 3.23. This characteristic curve is approximated (for the range to 
about 7 v) quite closely by the dashed line of Fig. 3.20. This dashed 
line represents a linear relationship between voltage and current. Since a 
resistor has the same linear relationship, the diode can be approximated 
(at least in the voltage range of to 7 v) by a resistor. In Fig. 3.23, 30 
milliamperes of current flow when 6 v is applied. Hence, the dashed line 
represents 6/0.03 = 200 £2. This dashed line resistance is known as the 
static plate resistance of the diode (at 30 ma.) and is given the symbol r P . 

If small voltage fluctuations (in the order of 1 v or so for Fig. 3.23) are 
encountered, a second type of equivalent resistance can be found. The 
change of diode voltage with diode current is the important relationship. 
Thus, if i P is the instantaneous value of plate current and v P is the instan- 
taneous value of plate voltage, 

r, = £* (3-65) 



92 

or in the limit, 



dvj. 
dip 



Electronic Engineering 



(3.66) 



where r v is known as the dynamic or small signal plate resistance. For 
example, if the plate of a 6AL5 tube is maintained near +6 v the dynamic 
plate resistance would have the slope given by the dotted line of Fig. 3.23. 



80 



60 



E 
■2 40 



20 

































Actual character 
curve 


stic 






















sy' 








Cu 


T/e of equivalent circ 
r p =200tt 


uit . 


" ^ 


' 






























^ 
^S* 


'Zr 














x' 
^ s 


^_S 




Curve of equivalent circuit 
"- r p =125Q 




^ — 


^-, 



















4 6 

Plate volts 



10 



Fig. 3.23. A method of finding the impedance to be used in an equivalent circuit for a 
6AL5 diode. 

Hence, a change of 1 v causes a change of 8 ma or r. = = 125 Q. 

6 " 0.008 

The circuit in which the diode is used will determine which type of equiva- 
lent plate resistance to use. However, most diode circuits will require the 
use of the static rather than the dynamic plate resistance. In contrast, 
the concept of dynamic plate resistance will be very useful in analyzing 
the devices to be considered in Chapter 4. 

Thus far, only part of the equivalent circuit for a diode has been found. 
The static plate resistance is only valid for the range to 7 v (the dynamic 
plate resistance is very accurate for plate voltages from 5 to 8 v). In the 
to 7 v range the equivalent circuit for the diode is as shown in Fig. 3.24a. 
If negative voltage is applied to the plate, essentially no current flows in the 
circuit, and the equivalent circuit for the tube becomes an open circuit as 
shown in Fig. 3.246. The total equivalent circuit for the diode is as shown 
in Fig. 3.24c. The switch 5 X of this figure is in position 1 if a positive 



Diodes 



93 




Plate 



"■Cathode 




(a) 


(b) 


(c) 


Forward bias 


Reverse bias 


Total 


equivalent circuit 


equivalent circuit 


equivalent circuit 



Fig. 3.24. Derivation of the equivalent circuit for a 6AL5 diode. The switch Si of part c 
is at position 1 for + voltage on plate and in position 2 for — voltage on plate. 

voltage is applied to the plate, and in position 2 if a negative voltage is 
applied to the plate. 15 

The semiconductor diode can be represented by an equivalent circuit as 
shown in Fig. 3.25. Again, the switch operates as explained previously. In 
this diode, the resistance in the forward direction r t is quite low (ohms or 
so) whereas the resistance in the reverse direction r b is quite high (megohm 
or so). The current vs voltage characteristics of the equivalent circuit are 
also shown in Fig. 3.25. The equivalent circuit shown applies only to 
diodes operating in the voltage region before breakdown occurs. Manu- 
facturers usually list the values of r f and r b for semiconductor diodes. In 
many cases diodes can be considered as simple switches which are on in 
the forward direction (77 ~ 0) and off in the reverse direction (/•„ ~ 00). A 
diode with this type of characteristics is known as an ideal diode. 



Very high 
resistance 



Positive electrode 




Si 
I Negative electrode 



(a) Equivalent circuit 



(b) Current vs. voltage characteristics 
of the equivalent circuit 



Fig. 3.25. Equivalent circuit of a semiconductor diode. 

15 The system of linear approximation given in Chapter 16 can be used with diodes to 
produce more accurate equivalent circuits. 



94 



Electronic Engineering 



The equivalent circuit for a gas diode and the current vs voltage charac- 
teristics of the equivalent circuit are shown in Fig. 3.26. Since the voltage is 
essentially constant over a wide range of current in the forward direction, 
this type of tube can be represented by a battery. When the bias is in the 
reverse direction, essentially no current flows so the equivalent circuit is an 



Add low value of 
R here for a 
zener diode 




Zener diode- 



Voltage - 



Cathode 



(a) Equivalent circuit 
Fig. 3.26. Equivalent circuit for a gas diode or for a zener diode. 



(6; Current vs. voltage characteristics 
of the equivalent circuit 



open circuit. Again, the switch S T operates as outlined previously. Figure 
3.26 can also be used to represent the zener diode in the constant voltage 
region of operation. Of course, the zener diode would not have high 
resistance if the bias voltage were reversed. 

In all the foregoing equivalent circuits, the impedances of the elements 
in the equivalent circuits are not affected by frequency. For low fre- 
quencies, the circuits are quite accurate. However, for very high fre- 
quencies (megahertz or higher) one additional element should be added 
to each circuit. As already noted for the semiconductor, a capacitance 
exists across the diode. Similar capacitances also exist between the plates 
and cathodes of gas and vacuum diodes. Consequently, each diode 
equivalent circuit should be shunted by a capacitor equal to the capacitance 
across the diode. These capacitances are usually in the micro-microfarad 
(or picofarad) range, so they only become effective at very high frequencies. 
The use of this capacitance is an oversimplification. A more accurate 
representation will be discussed in Chapter 16. 



3.15 USE OF DIODE EQUIVALENT CIRCUITS 

Some simple examples may help illustrate the usefulness of the equivalent 
circuits just devised. 



Diodes 



95 




i of a 5U4 



Bi = 2000Q- 



v = 100 sin 377t ( 



Fig. 3.27. Circuit for a half-wave rectifier. 



Example 3.2. One section of a 5U4 diode is to be connected as shown in Fig. 3.27. 
If the characteristic curve of the 5U4 diode is as shown in Fig. 3.28, find an 
expression for the current through the load, R L . 

To find an accurate value for r P , the peak voltage across the diode must be 
known. This peak value can be found by the graphical method of Section 3.13. 
In this example, the peak voltage applied to the circuit is 100 v and R L is 2000 CI. 
Therefore, the line AC (Fig. 3.28) represents the peak voltage conditions and 
point B indicates the maximum voltage across the diode. Since this maximum 
value of plate voltage is about 16 v, a resistance line is drawn which approximates 
the characteristic curve between and 16 v. This line (when extended) indicates 
100 ma flows when 40 v are applied. Hence r P is equal to 40/0.1 or 400 CI for 
the circuit given in Fig. 3.27. 

The two equivalent circuits shown in Fig. 3.29 approximate the action of the 



800 



600 



= 400 



200 



































Ac 


tual cu 


rve y 
































































































? 


= 400 


a 




c_ _ 


B^ti 
















A 



"0 20 40 60 80 100 

Plate volts 

Fig. 3.28. Characteristic curve of a 5U4 diode (one section). 



96 



Electronic Engineering 



R L = 2000a', 



ip\ 
#£ = 2000 0' 



(a) 

Equivalent circuit when 
sin 377t is negative 



(b) 

Equivalent circuit when 
sin 377* is positive 



Fig. 3.29. Equivalent circuit for Fig. 3.27. 

diode. Since sin 377/ is positive for / from to T f„ sec, the current i P can be 
found by the relationship 

100 

'p = r . „ sin 377/ (3.67) 

(iio > ' > 0) 

For this example, 100/(/> + R£) is equal to 0.042 amp. Also, since the sin 377/ 

is negative for / between T |, sec and ^ sec 

i P = (3.68) 

(xio <'<&) 

Now, since the sin 377/ is positive for / greater than ^ sec but less than T fo sec, 
Eq. 3.67 will also apply for this time interval. In fact, Eq. 3.67 applies for / > w/60 
but less than (2/i + 1)/120. By similar reasoning, Eq. 3.68 is valid for / greater 
than (2« + 1)/120 but less than (« + l)/60 where n is 0, 1, 2, 3, 4, and so forth. 

A plot of the current in R L can be made using Eq. 3.67 and Eq. 3.68. This 
plot is shown in Fig. 3.30. Again, the negative portion of the input signal has 
been clipped off. 

The average current can be found by integrating the total current in one cycle 





60 



_3_ 
120 




Time in seconds - 



Fig. 3.30. The current output of the circuit in Fig. 3.27. 



Diodes 



97 



and dividing by the time for one cycle. Therefore, 

[100/0p +R L )] sin 377r dt 






f 

Jo 



Kc 



or 






100 



(3.69) 



(3.70) 



(r P + R L )n 

The example just considered is known as a half-wave rectifier. The 
rectifier circuits are used in power supplies which convert a-c to d-c power. 
If R L is much larger than r P in Eq. 3.67, r P can often be ignored. When 
r P is ignored, the diode is assumed to be an "ideal diode." 

PROB. 3.21. A circuit is connected as shown in Fig. 3.27. Change R L to 
250 CI and plot i P . 

PROB. 3.22. One section of a 6AL5 (see Fig. 3.23) diode is to be connected as 
shown in Fig. 3.27. If R L is 20,000 CI, plot i P vs time. 

Many times the manufacturers list only forward resistance r f and reverse 
resistance r b for the junction diodes. Following is an example which 
illustrates how this information may be used. 

Example 3.3. A diode is connected in a circuit as shown in Fig. 3.31. The diode 
has a forward resistance of 100 CI and a reverse resistance of 100,000 CI. If a 
voltage with the waveform shown in Fig. 3.31 is applied to the input, plot the 
output voltage v Q . 




12 3 4 5 
Time in milliseconds 




Fig. 3.31. A diode clamper circuit. 

The capacitor in the circuit is assumed to be completely uncharged at the time 
t = 0. Then, at time t — 0+, the circuit would be as shown in Fig. 3.32a. This 
circuit is a simple series RC circuit whose loop equation can be written as 



10 = r,i + 



Solving this equation for /, 



X c\ idt 



i=— e -t/r/C 



(3.71) 



(3.72) 



98 



Electronic Engineering 



The term r f C is known as the time constant of the circuit and is usually given the 
symbol t. 
Since v is equal to i times r f , 

v = 10er" r (3.73) 

Thus at time / = 0, the output voltage is 10 v, but this output voltage decays 
rapidly (with a time constant of 10 /isec) to essentially zero. This portion of the 
output voltage is plotted (for 1 msec > t > 0) in Fig. 3.33. 




(a) Time t = 0+ 



"out 



10 v 
0.1 Mf 



Diode 



+ 



r b = 100 kfi 




(b) Time t = 1 millisecond + 




(c) Time t = 2 millisecond + 
Fig. 3.32. The equivalent circuits for Fig. 3.31. 

At time t = 1 msec, the capacitor is charged to essentially 10 v when the 
input voltage shifts to become - 10 v. Consequently, the equivalent circuit is as 
shown in Fig. 3.326, where 

v = -lOe-*' 1 ^ (3.74) 

where t' has the value of at the time t = 1 msec. Thus, the output voltage at 
time t = 1 msec is —20 v and decays toward potential. However, the time 
constant r b C is large (10 -2 sec) compared to one msec (the time during which this 
circuit is valid), so the capacitor does not discharge appreciably before the input 
voltage reverses again. Actually, 



v = -20e- 01 = -18.2 v 



(3.75) 



for the output voltage at the end of this portion of the cycle. This voltage 
variation is plotted in Fig. 3.33. 

Since the capacitor discharged only 1 .8 v during the period from the time t = 1 
msec to r = 2 msec, a charge of 8.2 v remains on the capacitor. Hence, at time 
t = 2 msec the equivalent circuit is as shown in Fig. 3.32c. In this case, 



v = l.Se-*"'" 



(3.76) 



Diodes 



99 



+ 10 



_t 



-10 - 



-20 



t= 10 Msec 

1.8 v 




2 3 4 5 

Time in milliseconds 

Fig. 3.33. A plot of v for Fig. 3.31. 



where t" has the value of zero at time t = 2 msec. As in the circuit in Fig. 3.32a, 
the capacitor quickly charges to 10 v again. From this point on, the action of the 
circuit is repetitious. 

The foregoing circuit is known as a clamper circuit. In effect, this 
circuit has "clamped" the most positive portion of the input waveform to a 
value of v. Many modifications of this circuit are used in various pulse 
circuit applications. 

PROB. 3.23. Repeat Example 3.3 if a diode with a forward resistance of 100 CI 
and a reverse resistance of 1 megohm is used. Does this diode improve the 
clamping action? 

PROB. 3.24. Plot the output voltage waveform if the diode in Fig. 3.31 is 
reversed in the circuit. Assume r f = 100 CI and r b = 100,000 CI. 

The circuit considered in Example 3.2 removed the negative portion of 
an a-c signal and was called a half-wave rectifier circuit. The circuit for 
a. full-wave rectifier is given in Fig. 3.34. 

|^ — dtl— 




Fig. 3.34. A full-wave rectifier circuit. 



100 



Electronic Engineering 



Example 3.4. A circuit is connected as shown in Fig. 3.34. The voltage input 
to the transformer T t is 100 v, 100 cycles, and the output voltage is 400 v peak on 
each side of the center tap. The diodes have low forward resistance and very 
high back resistance. The capacitor has a capacitance of 10 //f and the load 
resistor has a resistance of 1000 CI. Plot the voltage across R L as a function of 
time. 

Let time t = when the terminal 1 of transformer 7\ is at the maximum 
positive potential, Since diode D 2 has a reverse voltage, very little current will 



400 
395 



278 



^Exponential decay 




Fig. 3.35. Output voltage waveform of the circuit of Fig. 3.34. 



flow through side 3 of the transformer. In fact, since the reverse current is so 
small, the reverse current can be ignored without introducing any appreciable 
error. Since side 1 is positive with respect to ground, a current will flow through 
diode Dj_. This current will charge capacitor C as shown in Fig. 3.35 (curve A to 
B) and also cause current to flow through R L . If the forward resistance of the 
diode and the resistance of the transformer are ignored, the capacitor will charge 
to the peak value of the voltage (400 v) across 1/2 of the transformer secondary. 
After the peak value of voltage is past, the voltage at point 1 will decrease as 
shown by the dashed line B to C of Fig. 3.35. As soon as the voltage at point 1 
becomes less than the charge on the capacitor, the diode D x is biased in the 
reverse direction and the diode becomes an open circuit. The charge on the 
capacitor C starts to leak off through the resistor R L as soon as D 1 is cut off. 

The performance of this circuit can be analyzed mathematically. Thus, the 
voltage across the capacitor, v c between time t± and r 2 is given by the equation 

v c = y ie -(t-t,)IXLC (3 77) 

Ci < / < h) 

where V x is the voltage at which diode D 1 ceases to conduct, and t 1 is the time 
when this cut-off of diode D 1 occurs. Also, the voltage on the transformer side 
of the diodes, v T (curve ABCDEF, etc., in Fig. 3.35) is given by the equation 



V T = \V m cos ODt\ 



(3.78) 



Diodes 101 

where V m is the peak value of transformer voltage and to is the radian frequency 
of the power source. 

Inspection of Fig. 3.35 indicates that V x occurs when the slope of v c is equal to 
the slope of vt. 

d ^£ = _ Jj_ e -u-tjlR L c (3 79) 

dt R L C 

and 

dVrn 

-j- = -<aV m sm<»t (3.80) 

Now, these two derivatives will be equal when t = t x . Hence, 

-mV m taaut 1 = --=-7; (3.81) 

The voltage, V lt is equal to vt at the time t x . Therefore, 

V x = V m cos mt x (3.82) 

When Eq. 3.82 is substituted into Eq. 3.81, 

V m COS CO?! 

coV m sin a>/ x = =-— — (3.83) 

or 

tan ^ - ^ (384) 

This equation can be written as 

at! -tan" 1 — — (3.85) 

ct)R L C 

or 

,^-tan- 1 — J— (3.86) 

If the value of ^ from Eq. 3.86 is substituted into Eq. 3.82, 

K^^cos^an-^) (3 - 87) 

This equation can be simplified by referring to Fig. 3.36. If 6 = tan -1 \j(oR L C, 
has the value shown.. Hence, the cos is (oR L Cj[\ + {(oR L CfV A and Eq. 3.87 
becomes 

V m a>R L C 

Ki = [x +{W R L cm (388) .^M^ 




The value of V 2 (the voltage when diode 
D 2 begins to conduct) can be found by "^ ' u r l c 

equating v c and v? at the time / = t 2 . 

Accordingly, when the value of V x from Fig. 3.36. A graphical representation 
Eq. 3.88 is substituted into Eq. 3.77 and of Eq. 3.91. 



102 Electronic Engineering 

when Eq. 3.77 is equated to Eq. 3.78, 

V m o>RjC 

1 v - c ° s ■*•" - [i+r^o^ e " h ~ ti)iRtc (3 - 89) 

or 

,cos ^ - [i + sro^ e - {H - tmLc **» 

Since f x can be found from Eq. 3.86, and the values of m, R L , and C are known 
for a given circuit, the time t 2 is the only unknown in Eq. 3.90. Unfortunately, 
Eq. 3.90 cannot be solved directly and therefore must be solved either graphically 
or by trial and error. 
For this example, 

coR L C = 628 x 10 3 x 10- 5 = 6.28 (3.91) 

tan_1 T^E = 9° = 0.157 radian (3.92) 

6.28 

1 
h = 7ZZ x 0-157 = 2.5 x 10-*sec (3.93) 

ozo 

400 x 6.28 
^ (l + 39.4)H = 395v < 3 - 94) 

t 2 by trial and error =* 3.75 x 10~~ 3 sec 
K 2 = 278 v 

As mentioned previously, the current through each diode flows only during 
a short period of time each cycle. While the current flows, it has the value 

i = c ir + f L < 3 - 95 > 

The term C(dv T jdt) represents the current used to charge the capacitor while 
the current v T jR L is the current delivered to the load. Usually the current 
required to charge the capacitor is many times greater than the current required 
by the load resistor. Neglecting the source resistance and inductance, the 
maximum current flow through the diode usually occurs at t 2 when dv T \dt is 
greatest. Hence, a quick check allows the circuit designer to determine if the 
current through the diode exceeds the recommended maximum. 
Substituting Eq. 3.80 and Eq. 3.78 into Eq. 3.95, 

V m cos cof 2 
'max at t 2 = CmV m sin cot 2 + (3.96) 

K L 

= 10~ 5 x 628 x 400 sin (628 x 3.75 x lO" 3 ) 
400 cos (628 x 3.75 x 10" 3 ) 
+ 1000 

= 1.77 + 0.283 = 2.053 amp 



Diodes 



103 



Since actual circuits always contain some inductance and resistance, / max occurs 
later than t 2 and is less than that given by Eq. 3.96. 

A derivation similar to the foregoing can be made for the half-wave rectifier. 
In the half-wave rectifier, the time when conduction ceases (t t ) is the same as 
in the full-wave rectifier. However, the time when conduction begins (t 2 ) will 
be different. A plot is made in Fig. 3.37 to help determine the value of t x and 
t a for either the full-wave or half-wave rectifier. 




r-i CM ^ ID ■-< N ^IO o O OQQ Q 
cSooo .-tCM^'DOO 



uR L C 
Fig. 3.37. A plot to help determine t t and /, in Fig. 3.35. 

PROB. 3.25. Repeat Example 3.4 if C is changed to 100 pi. What is the peak 
current through the diode. 

PROB. 3.26. The diode D x (in Example 3.4) has a failure and becomes an open 
circuit. Plot the output voltage waveform. What is the peak current through 



The foregoing derivation of relationships in a rectifier-filter system is 
adequately rigorous, but disappointingly tedious from the standpoint of 
system design. Considerable simplification results when certain approxi- 
mations, which will be discussed, are allowable. Whenever a capacito r 
is used in a rectifier system t o improve the rectification efficiency and 
r educe the a-c component or ripple in the load, the capacitance value is 
usually chosen so that the ripple voltage is small in comparison with the 
d-c component of load voltage . Under this condition, the time constan t 
of the load resistance and filter capacitance must be long compared with 
t he period 7" of the input voltage (Fig . 3.38). Then the capacitor (and load) 
voltage decreases almost linearly at the initialdischarge rate V m *JR T .C 
(obtained from Eq. 3.79 by letting t — /, and assuming t>, = p,^). This initial 
slope would reduce the load voltage to zero at f = R L C if it were allowed to 
continue. 



104 



Electronic Engineering 



T 7 Vmax 


A B' 

\ 1 / 1 


1 1 ( 


1 1 1 / 1 


B 


K min 


\ ■ / 1 1 r-^-^l 



R L C 



Fig. 3.38. Constructions used in the approximate solution. 



If semiconductor diodes are used as rectifiers, the maximum voltage 
across the load is approximately equal to the peak input voltage, since 
the forward drop across the diode is approximately one volt. Then an 
approximate relationship between the filter capacitance and the ripple 
voltage can easily be obtained by the following procedure. 

1. Assume that the load voltage decreases linearly from t = until t = T, and 
then the capacitor is instantly recharged to Kma* and so on. Then triangle 
AB'D' is similar to triangle ABD and 

-ZL-£51 (3.97) 

Rj^C Kmax 

2. B'D' is the peak-to-peak ripple voltage which may be approximated from 
the specified rms ripple voltage. B'D' ~ 2 V2 v rlpp i e . 

3. The average or d-c load voltage is obtained by subtracting the peak ripple 
voltage B'D'jl from Kmax, or, more often, the maximum input voltage is 
obtained by adding the peak ripple voltage to the specified d-c load voltage. 

4. The minimum value of load resistance, which is the worst case, may be 
obtained by dividing the d-c load voltage by the maximum specified load 
current. 

5. The required value of capaciianc&can be,pbtained from the specifications for 
the power supply and with the aid of Eq. 3.97 

C = J Vm * x (3.98) 

2v2v rl p ple .R L 

When a full-wave rectifier is used, the discharge period is approximately 
7/2 instead of T. Therefore, the required filter capacitance is reduced 
by a factor of two. 

PROB. 3.27. Design a full-wave power supply with a capacitor filter which 
will provide 50 volts d-c at 1 ampere into a resistive load. The permissible rms 
ripple voltage is 2 per cent of the d-c load voltage. Use silicon rectifiers and 
determine the filter capacitance as well as the rms voltage of the transformer 
secondary. The primary power is 1 1 5 v 60 Hz. 

A computer solution of the equations in Example 3.4 is presented in 
nomograph form in Fig 3.39. This nomograph is very useful when deter- 
mining the ripple factor as a fraction of a>R L C. 



Diodes 



105 



mill i — i — i i i i i i i i — r 




sunjo ui 7 y 




105 Electronic Engineering 

Hot cathode gas diode tubes are frequently used as rectifiers. The 
voltage drop across these tubes is small in comparison with a vacuum 
diode (12 v is a typical value). Consequently, gas tubes have lower internal 
losses at high currents. In contrast, cold cathode gas diodes are usually 
used as voltage regulators. Of course, the zener diode is also used as a 
voltage regulator. The action of a voltage regulator can be demonstrated 
by an example. 

Example 3.5. A gas diode (type 0B2) has the following characteristics. Operat- 
ing voltage is 108 v for a current of 5 to 30 ma. The diode is connected in a circuit 
as shown in Fig. 3.40. What values may R L have in order for the voltage across 
R L to remain 108 v? 

+3000 <AMAA 

5K 



0vo 



Fig. 3.40. A voltage regulator circuit. 

The voltage drop across the 5 Kfi resistor must be 300 - 108 = 192 v. There- 
fore, the current through this resistor must be 

192 
/ = — = 38.4 ma 

The current through the diode can vary from 5 to 30 ma and still maintain 108 v 
across the diode. Therefore, the current through the load can change between 
8.4and33.4ma. Hence, the values of load resistance will be 10 %.oo84 = 12,9000 
tO 108 / .0334 =3240 Q. 

PROB. 3.28. The diode of Example 3.5 is connected as shown in Fig. 3.40. If 
R L is 5000 ii, over what voltage range can the 300-v supply vary and still 
maintain 108 v across R L 1 Answer: Vmm = 241 v, Kma* = 366 v. 
PROB. 3.29. A circuit is connected as shown in Fig. 3.416. Plot the output 
voltage waveform if the input voltage has the form shown in Fig. 3.41a and the 
characteristics of the diode are as shown in Fig. 3.41c. 

PROB. 3.30. A circuit is connected as shown in Fig. 3.42. Plot the steady-state 
output waveform if the input signal has the form shown. 
PROB. 3.31. A junction diode has 1 v reverse bias applied. The measured 
current is 0.1 fin. What current would flow (T = 300°K) if 10 v forward bias 
is applied? The internal resistance of the diode is 0.25 ft. 
PROB. 3.32. A germanium diode (at room temperature) has doping con- 
centrations such that N a = 10 17 /cm 3 and N d = 10 15 /cm 3 . 



Diodes 

+25 v 
+ 10 



+25 Vj— , 
.Ovr- 1 



Ovl— . 
-25 v I— 



-lOvl-, l_ 

-25 v I 
(a) 



i 2.5 K "out 

, i 



(b) 



+ 10 



s. 

i +5 



-30 -20 -10 



107 



10 20 

Volts 



-5 



-10 



30 



(c) 

Fig. 3.41. Information required by Prob. 3.29. (a) Input voltage; (b) circuit configura- 
tion; (c) characteristic curves. . 



1 


M 

iy = 100O 






f 

"out 


'in 


h=iook 




ZlfM 


I 








1 



-0.1 milliseconds 



+ 10v 



-10 millisec- 
Fig. 3.42. Information for Prob. 3.30. 



108 Electronic Engineering 

(a) What are the values of n„ and/?„? 

(6) What are the values of w„ and /?„? 

(c) What is the magnitude of the potential K ft0 ? 

(a") What are the lengths of L v and Z,„? 

(e) What is maximum electric field intensity in the depletion region? 

PROB. 3.33. A silicon diode (at room temperature) has the following physical 
parameters: 

N a = 10 17 /cm 3 

N d = 10 15 /cm 3 

Cross-sectional area = 0.01 cm 2 

(a) Determine the junction capacitance of this diode when no external voltage 
is applied. 

(b) Make a plot of capacitance vs external voltage for this diode if the ex- 
ternal voltage were to vary from —25 to v. 

PROB. 3.34. Write a computer program which will read-in values of o>, R L , C, 
and V m . The computer should solve the equations given in Example 3.4 and 
print (or punch) out the values of t lt t z , V lt V 2 , and 1^ ve . 



4 



Basic Amplifiers 



The vast electronics industry is centered around the amplifier, a device 
capable of increasing the power level of an input signal. For example, the 
minute amounts of power collected by a radio-receiving antenna are 
entirely inadequate to operate a loudspeaker or provide a picture on a TV 
picture tube. Amplifiers and other electronic circuits are required in 
order to increase the power level of the received signals and make them 
suitable for operating the output devices. In addition to its major role in 
the communications area, the amplifier is a basic building block in 
automation and computer industries. Also, electronic instruments that 
employ amplifiers have become an essential part of the medical profession. 

The amplifier is actually an energy converter. The input signal merely 
controls the current that flows from the power supply or battery. Thus, the 
energy from the power supply is converted by the amplifier to Signal 
energy. There are many types of amplifiers in current use, including 
hydraulic, pneumatic, magnetic, transistor, and vacuum tube amplifiers. 
Only part of the transistor and vacuum tube types will be included in this 
treatment. The basic principles of almost all types are similar, however, 
so a thorough understanding of one will give insight into all. 



109 



110 



Electronic Engineering 



4.1 JUNCTION TRANSISTORS 

When a semiconductor is arranged so that it has two p-n junctions as 
shown in Fig. 4. la, it is known as ap-n-p junction transistor. The electrical 
potential in the transistor as a function of distance along the axis is 
sketched in Fig. 4.1b. If forward bias is applied to one junction and 
reverse bias is applied to the other junction, as shown in Fig. 4.1c, the 
potential diagram will be altered as indicated in Fig. 4.ld. The reduction 
in the height of the potential barrier of the forward biased junction will 
cause positive carriers from the left-hand region, which is known as the 
emitter, to be injected into the center section, which is known as the base. 
Some of the injected holes recombine with the negative carriers in the base 
region, but if the base region is made very thin, the majority of the injected 



Emitter >. Base 




1 ' I 

' II I 

(a) p-n-p Transistor 

I ■■' ' ' 



I 



(b) Idealized potential of (a) 




, iii 



Collector 



«c" 



<c) 


Biased 


p-n-p transistor 

1 


75 


1 




1 
1 


c 


Veb 








Q_ 


t 




V CB 

\ 






(e) p-n-p Transistor circuit 



(d) Idealized potential of (c) 
E C 




(f) n-p-n Transistor circuit 



Fig. 4.1. The p-n-p junction transistor, (a) p-n-p transistor; (b) idealized potential of 
a; (c) biased p-n-p transistor; (d) idealized potential of c. 



Basic Amplifiers 



111 



holes diffuse across the base and are accelerated through the reverse biased 
junction into the right-hand section, which is known as the collector. The 
holes drift across the collector region and cause current i c to flow in the 
collector circuit. Also, a base current i B flows as a result of recombinations 
in the base. A circuit diagram of the p-n-p transistor and bias batteries is 
given in Fig. 4.1e. The transistor could have been an n-p-n arrangement, 
in which all potentials and currents would be reversed. A circuit diagram 
including bias batteries and the transistor symbol for an n-p-n transistor is 
given in Fig. 4.1/. 

From the preceding discussion, it would seem that either end of the 
transistor could be used as the emitter. This conclusion is generally not 




V BB ■%• Vcc 

(a) (b) 

Fig. 4.2. Common-base transistor amplifiers. 

true. The heat dissipation at the collector junction is much greater than 
the dissipation at the emitter junction because of the comparatively large 
potential difference across the collector junction. Therefore, the transistor 
is usually designed so that the heat can be most effectively conducted away 
from the collector junction. The power dissipation rating of the transistor 
depends on the thermal conductivity from the collector junction to the 
transistor case and thence to the ambient surroundings. 

At this point, a few words should be said about conventional current 
and voltage directions. Actual current directions and potential polarities 
were shown in Fig. 4.1. However, IEEE standards require that currents 
which flow into the device be considered positive, and that currents which 
flow out of the device be considered negative (Fig. 4.2). Therefore, the 
collector and base currents which usually flow out of the p-n-p transistor 
are negative currents indicating that they flow in the opposite direction to 
the conventionial positive currents. Also, a potential is considered positive 
if it is positive with respect to common node. The arrow head points in 
the direction of positive potential, as shown in Fig. 4.2. In the p-n-p tran- 
sistor, the collector is actually negative with respect to the base, which 
is the common node in the common base configuration. Therefore, the 
actual p-n-p collector voltage is negative. 



\ 



b 



112 Electronic Engineering 

A 

When the voltage between the collector and the base is held constant, 
as in the preceding discussion, the fraction of the emitter current which 
reaches the collector terminal is called a. In addition to this current, the 
saturation current resulting from thermally generated carriers flows across 
the reverse biased collector-base junction, as discussed in Chapter 3. This 
saturation current was known as I 8 in the diode but is known as I co or I CBO 
in the transistor. 1 Since I co flows out of the collector in a p-n-p tran- 
sistor, I co will be a negative current. Then the collector current is 

—i c = cd E - l co (4.1) 

(Generally! tD£ couector circuit can be thought of as the output circuit. It is 
therefore desirable to have as large a collector current as possible, or in 
other words, to have a approach unity. This can be accomplished by 
making the base, section thin) as previously mentioned, to reduce the 
opportunities for recombination. Also, the doping concentration in the 
base region may be a minimum practical value in order to provide a 
minimum recombination rate. In currently available junction transistors, 
values of a range from about 0.95 to 0.998. 

Using KirchhofT's current law, it is evident from Fig. 4.1 that the base 
current is 

— (b = h + *c ( 4 - 2 ) 

—h = h — (*'« — ho) = - a )'is + I co ( 4 - 3 ) 

PROB. 4.1. A certain transistor has a. — 0.98. V BE is adjusted so that i E = 10 
ma. Assuming that the collector junction is reverse biased and I co = —10 #a, 
calculate i c and i B . Answer: i B = 190 pa, i c = 9.82 ma. 
PROB. 4.2. At what magnitude of emitter current will the base current be equal 
to zero? 



4.2 THE COMMON-BASE AMPLIFIER 

If the emitter and base terminals are the input terminals and the collector 
and base terminals are the output terminals, the transistor is in the common 
base or grounded base configuration. A gain in signal power may be 
realized with this arrangement. The j nput re sistance of this configuration 
is low because of the forward bias on the emitter-base junction. In contrast, 

1 The saturation current is frequently known as I BB0 in the literature. The three 
subscripts indicate this current flows from the collector (first subscript) to the base 
(second subscript) when the emitter circuit is open (the third subscript indicates the 
condition of the third terminal of the device). For example, when the emitter junction 
is reversed biased and the collector circuit is open the current which flows is known as 
I XB0 . Since the second subscript is always B when thermal currents are involved, this 
second subscript may be eliminated without loss of information. 



Basic Amplifiers 113 

the o utput resistance is hi gh because of thejeverse bias j_ onjthe collector- 
basejunction. Therefore, a load resistance which is high in comparison 
wltritheinput resistance may be placed in the collector circuit. Since the 
output current may be almost equal to the input current, the ratio of power 
in the load resistance to the power input (known as power gain) may be 
almost equal to the ratio of load resistance to input resistance. If the input 
and load currents are almost equal, the voltage gain must be essentially 
equal to the power gain. When there is no means provided to separate the 
bias currents from the signal currents, the amplifier is said to be a d-c 
amplifier. A schematic diagram for a common base d-c amplifier is 
given in Fig. 4.2a. 

Some practical problems may arise from the lack of common d-c 
potentials in the input and output circuits of this amplifier. These 
problems are avoided in the circuit of Fig. 4.2b. This circuit utilizes 
capacitors to isolate the bias currents from the signal currents. This 
amplifier is called an a-c amplifier. In this amplifier the bias supply 
voltage V BB must be increased in order to compensate for the drop 
across R x . The value of R t should be large in comparison with the input 
resistance of the transistor so the signal current in the emitter of the 
transistor is almost equal to the source current i ( . In either the d-c or 
the a-c circuit, a small variation in the input voltage causes a change in the 
height of the emitter-base junction potential barrier, which in turn causes a 
significant change in the injection current across this barrier. Most of the 
injected carriers drift across the narrow base region into the collector and 
thus cause collector-current variations which are almost equal to the 
emitter-current variations. The collector-current variations cause voltage 
variations across the load resistance R L . These voltage variations may be 
much larger than the input voltage, v t . The voltage variations across R L 
constitute the output signal voltage, which may be isolated from the bias 
voltage by the capacitor C 2 . Observe that the varying or signal com- 
ponents of current or voltage are the total values minus the bias values. 
Notice also that the bias potentials applied to the transistor, V EB and 
V CB are different from the bias battery potentials V BB and V cc . 

When the load resistance is small in comparison with the transistor 
output resistance, /„ is approximately equal to xi e . This statement will be 
verified later. Then, neglecting the shunting effect of R u we have 

»i = Win ( 4 - 4 ) 

(4.5) 
The voltage gain is 

(4.6) 



v„ 




-i c R L ^ 


a 'e*£ 


G v = 









114 Electronic Engineering 

The instantaneous power input is 

Pi = W« (4.7) 

The instantaneous power output is 

Po = i*R L ~(xi e yR L (4.8) 



The power gain is 



G„ = - - — — — = — — (4.9) 

Pi l e K in K in 



The foregoing relationships are based on the following assumptions: 

1. The input current is proportional to the input voltage, so that R in may be 
defined. 

2. The load resistance R L is small in comparison with the output resistance of 
the transistor. It is evident that the output current i c must approach zero as 
the load resistance approaches infinity. 

3. The frequency of the signal components is such that all reactances in the 
circuit are negligible. 

PROB. 4.3. A certain transistor has an input resistance of 50 ii, an output 
resistance of 1 megohm, and an <x of 0.98. Calculate the voltage and power gains 
in a common-base circuit which has a load resistance of 10 Kii. Assume that the 
conditions listed are fulfilled. Answer: G v = 196, G v = 192. 

4.3 GRAPHICAL ANALYSIS 

In order to use a transistor intelligently, some basic relationships 
between its currents and voltages must be known. For example, the 
relationship between the emitter-base voltage and the emitter current 
might be very helpful. Also, the effects of emitter current and collector 
voltage on collector current would be very useful. These relationships 
could be calculated from theoretical considerations if sufficient information 
were supplied by the manufacturer. However, it is much easier and more 
accurate to determine these relationships experimentally. The diagram of a 
circuit which might be used to obtain the common-base characteristics is 
given in Fig. 4.3. The batteries and potentiometers which are used as 
variable bias sources in this circuit are customarily replaced by a-c 
operated power supplies with adjustable d-c output voltages. The emitter 
bias voltage would be low (a few tenths of a volt) and its adjustment 
would be rather critical if the resistor R were not included in the circuit. 
The voltmeters shown should be highly sensitive, preferably vacuum tube 
voltmeters, so that their operating currents do not make an appreciable 
contribution to i c or i E . 



Basic Amplifiers 



115 




-vww- 



© 



■WVW 1 



v BB Vcc 

Fig. 4.3. A circuit for determining transistor characteristics. 

There are four important variables in the circuit, as indicated by the 
meters. Some of these variables must be held constant while the relation- 
ships between others are found. For example, the emitter current must be 
held constant while the effect of the collector voltage on collector current is 
determined. When the collector current is plotted as a function of the 
collector-base voltage (with emitter current held constant), the resulting 
curve is known as a collector-characteristic curve. Each different value of 
emitter current will yield a different collector-characteristic curve. A set of 
several curves obtained from several representative values of emitter 
current is known as a family of collector-characteristic curves. A typical 
family or set of collector characteristics for a p-n-p transistor is shown in 
Fig. 4.4. The negative current indicates that the current is flowing out of 
the transistor, where the reference direction is into the transistor. A set of 

-5 



= -A 
2L 4 



1 -3 



-2 



I- 

















r 








5 ma 






s 








4 ma 




// 


s 








3 ma 






<- 








2 ma 




I 


•■ 






if = 1 ma 





+ 1 -5 -10 -15 -20 -25 -30 

Collector to base voltage in volts 
Fig. 4.4. A family of collector characteristic curves. 



116 



Electronic Engineering 



curves for an n-p-n transistor might be identical to the set shown except 
that the polarities of currents and voltages would be reversed. 
From Fig. 4.4 it should be observed that : 

1. The collector current is almost equal to the emitter current when reverse bias 
is applied to the collector junction. 

2. The collector current is almost independent of collector voltage when reverse 
bias is applied to the collector junction. 

3. The collector current is rapidly reduced to zero and then reversed when 
increasing forward bias is applied to the collector junction. This behavior 
occurs because the injection current across the collector junction opposes the 
injection current across the emitter junction. The collector current is 
essentially the algebraic sum of the two injection currents. 

The emitter-base voltage does not appear in the set of collector charac- 
teristics. Therefore, a relationship is needed between emitter current and 
emitter-base voltage to determine the input resistance of the transistor. 
Experimentation reveals that the collector voltage has a slight influence on 
the emitter current. Therefore, a family of curves is needed to completely 
define the input characteristics of the transistor. Normally the emitter-base 
voltage is plotted as a function of the emitter current with the collector 
voltage held constant (Fig. 4.5). A curve is obtained for each different 
value of collector voltage. 

The characteristic curves just described are known as static-characteristic 
curves because certain parameters were held constant while the relationships 



0.8 



S 0.6 



§ 

o 



•8 0.4 



E 



0.2 

















*0 


-■£ r- 

























2 3 4 

Emitter current i s milliamperes 



Fig. 4.5. Input characteristics of a typical transistor. 



Basic Amplifiers 117 

between other parameters were obtained. When a transistor is used as an 
amplifier, there must be a load impedance in the collector circuit to 
produce an output voltage. The variations in input or emitter current and 
the resulting variations in collector voltage represent the signal compo- 
nents in the circuit. The characteristics of an amplifier in which all 
terminal voltages and currents are allowed to vary simultaneously are 
known as dynamic characteristics. The dynamic characteristics could be 
obtained by placing a load resistance in the collector circuit and plotting 
a set of curves for each value of load resistance. However, this method 
would become very tedious if many values of load resistance were 
considered. A much easier method of determining the dynamic charac- 
teristics utilizes the static characteristic curves which are either furnished 
by the manufacturer or easily obtained in the laboratory. This method is 
developed as follows. 

When a load resistance is placed in the collector circuit, as in Fig. 4.2, 
the collector-base voltage v CB at any instant is the collector supply voltage 
V cc minus the drop across the load resistor. Then, letting the voltages 
and currents carry their appropriate signs, 

(4.10) 
(4.11) 



Solving for i c 


v cb — V cc — ic R L 


or 


Vcc ~ v cn 
is* — 




f V CB + VCC 

° Rl Rl 





(4.12) 

Compare Eq. 4.12 with the familiar equation of a straight line 

y = mx + b (4.13) 

where m — the slope of the line 
b = the y axis intercept 

It is evident that Eq. 4.12 is the equation of a straight line which has a 
slope of — 1 /R L and i c axis intercept of V cc jR L . Equation 4.12 shows that 
when i c = 0, v CB = V cc , or in other words, the v CB axis intercept is K cc . 
This line is known as the load line. When drawn on the static collector 
characteristics of a transistor, the load line gives the dynamic operating 
characteristics for any chosen value of load resistance. A typical family of 
collector characteristics is provided in Fig. 4.6a. The load line is drawn for 
a 5 Kfi load resistance with V cc = 25 v. A bias point or quiescent point 
must be selected some place along the load line. If the expected input 



118 



Electronic Engineering 




o 
a 
.2 

2 

o 
u 




•o 

a 

ID 



V 






eiu 3 » 



Basic Amplifiers 119 

signals are symmetrical about the quiescent or bias value of i c , it would 
be wise to select the quiescent or bias point at about the center of the load 
line. This would be the case for sinusoidal input signals. On the other 
hand, if a positive pulse were to be applied to the input, the quiescent point 
preferably might be located at the lower end of the load line. 

The emitter-base voltage, for any combination of emitter current and 
collector voltage, may be found from the input characteristic curves, as 
previously discussed. Therefore, a dynamic operating line may be drawn 
on the input characteristics using the combinations of emitter current and 
collector voltage found along the load line on the collector characteristics. 
This dynamic input characteristic is illustrated in Fig. 4.6b. 

The dynamic operation of the transistor amplifier can be visualized with 
the aid of the curves of Fig. 4.6. In this sketch, the emitter current is 
assumed to be varying sinusoidally about the chosen quiescent operating 
point with maximum amplitude I e max = 1 ma as shown. Then, the 
amplitude and waveform of the collector-current variations are projected 
on the collector current axis as shown. Similarly, the collector-voltage 
variations are projected on the collector-voltage axis as indicated. Note 
that time axes have been drawn normal to the load line, current axis, and 
voltage axis, respectively. From the construction in Fig. 4.6a, it may be 
observed that the collector current and voltage have the same waveform 
as the emitter current, providing that the collector current is proportional 
to the emitter current, or in other words, if a is constant. 

The magnitude and waveform of v EB may be obtained from the input 
characteristics by projection of the assumed excursions of emitter current 
along the dynamic input characteristic to the v EB axis as shown in Fig. 
4.66. Since the dynamic input characteristic is not straight, the waveform 
of v EB is not the same as the waveform of i E , or in other words, i E is not 
proportional to v EB . This behavior is to be expected from the preceding 
discussion of the properties of a p-n junction. The waveform of v EB 
could be plotted quite accurately if we consider enough points at equally 
spaced time or degree intervals along the i E curve and project them from 
the dynamic curve to produce the v EB vs time plot. This detail is not 
warranted at this time, however. The change in waveform is known as 
nonlinear distortion and will be treated in detail in Chapter 10. 

It may seem more reasonable to assume that the input voltage v EB is 
sinusoidal and the input current i E is distorted rather than vice-versa. 
This condition would be true if the signal source had an internal impedance 
small in comparison with the input resistance of the transistor. Then the 
transistor input voltage would be almost equal in magnitude and hence 
similar in waveform to the driving source emf. The emitter current and 
consequently the collector current and voltage would then have nonlinear 



120 Electronic Engineering 

distortion. However, if the driving source has an internal impedance 
which is large in comparison with the transistor input resistance, the 
variations in input resistance are very small in comparison with the total 
impedance in series with the driving emf (visualizing a Thevenin's equiva- 
lent circuit) and the input current has essentially the same waveform as the 
source emf. The transistor input voltage would then be distorted as in the 
example given in Fig. 4.6, but the output voltage would be essentially 
undistorted. It may be observed from Fig. 4.6b that the used segment of 
the dynamic input characteristic approaches a straight line as the emitter 
current variations are reduced in magnitude. Thus the distortion may be 
negligible under any conditions of operation if the signal amplitude is 
sufficiently small. 

The voltage gain, current gain, and power gain of the amplifier may be 
obtained from the curves of Fig. 4.6 by a comparison of the relative 
magnitudes of the output excursions of current and voltage with those of 
the input. Comparison of peak-to-peak amplitudes yields the most 
accurate results because of the lack of symmetry of some of the waveforms. 
The power gain is the product of the voltage gain and the current gain, 
since the input impedance and load impedance are assumed to be resistive. 

PROB. 4.4. Determine the voltage gain, current gain, and power gain of the 
transistor amplifier represented in Fig. 4.6. Answer: G v = 80, G t = 0.98, 
G P = 78.5. 

PROB. 4.5. Using the characteristic curves of Fig. 4.6, draw a load line for 
R L = 10 Kfl and V cc = 20 v. Assuming that the emitter signal current will be 
0.5 sin <ot ma, select a suitable quiescent operating point and draw the dynamic 
input characteristic. Sketch the emitter-base voltage, collector voltage, and 
collector current and calculate the voltage gain, current gain, and power gain . 



4.4 FACTORS WHICH AFFECT « 

In the preceding discussion it has been implied that the current ratio a 
varies from one transistor to another but is constant for a particular 
transistor. This is not strictly true. The dimensions of the transistor as 
well as the doping concentrations, which are built into the transistor, do 
primarily determine a as previously mentioned, but the magnitude of 
applied voltages and currents also have some effect on a. The importance 
of this parameter a may not be appreciated at this point in the discussion 
of transistor amplifiers, but as the theme unfolds a will emerge as one of 
the chief characters. Therefore, an investigation will be made to determine 
what factors affect a. 

Figure 4.7a is a sketch of the cross section of a typical alloyed junction 
transistor. The base section is a thin wafer a thousandth of an inch or so 



Basic Amplifiers 



121 



thick. The emitter and collector are small beads of doped material alloyed 
to the base. This sketch is much larger than an actual transistor. Other 
methods of manufacture, which produce physical structures quite different 
from this one, are in use. The physical construction for the various types 
of transistors is usually described in the transistor manuals. 

The current flow through the base region is primarily by diffusion. The 
main objective is to have as few charge carriers as possible lost by recom- 
bination. Assuming the transistor to be ap-n-p type, the holes which are 



r^ 



Emitter/ |\ |— * 
Base-/ 



"\ 



Collector 



2 * 

<B ui 
t3 = 
5J o» 

= ' 




fa) 



(b) 



Fig. 4.7. Properties of an alloyed junction transistor (the base width is highly 
exaggerated). 

injected into the base diffuse in a radial fashion as indicated in the figure. 
Those carriers which reach the collector junction are swept into the 
collector and contribute to the collector current. The carriers which strike 
the outside perimeter of the base will almost certainly recombine because 
of the deathnium centers or traps existing at the surface of the material. It 
is evident from the figure that the percentage of carriers which find their 
way to the perimeter is reduced as the base thickness is reduced. Also, the 
recombinations in the body of the base region may be made small by 
providing low doping concentrations and by making the base thin. The 
base thickness (or axial length) must be small in comparison with the 
diffusion length. The diffusion length is defined as the distance the carriers 
will travel before their number is reduced by recombination to p je, 



122 Electronic Engineering 

where p is the initial concentration and e is the base of the natural 
logarithm. 

The ratio of the current which arrives at the collector junction to the 
initial current resulting from the carriers injected into the base is known 
as the transport factor, y. This transport factor depends primarily on the 
thickness of the base and is always less than unity. The effective base 
region is that portion of the base in which there is essentially no electric 
field. This effective base width, therefore, excludes the depletion region or 
potential hill at each junction. Because of the small potential difference 
across the emitter-base junction, the depletion region of this junction is 
very thin and may be considered essentially zero. The thickness of 
the depletion region is proportional to the square root of the potential 
difference across the junction (Eqs. 3.39 and 3.40). In addition, the penetra- 
tion is inversely proportional to the square root of doping concentration 
(Eqs. 3.39 and 3.40). Therefore, the depletion regions are primarily in the 
base material since the base is lightly doped in comparison with the emitter 
and collector. Thus, the effective base width, indicated by w in Fig. 4.76, 
decreases as the collector voltage is increased. The penetration into the 
base region of the collector junction depletion region is indicated by the 
cross-hatched area of Fig. 4.7b. Thus the transport factor increases as the 
collector voltage is increased. 

If the collector voltage is increased sufficiently, the effective base width 
w may become zero. This condition is known as punch through. Then, 
the transistor behaves as though the collector were connected directly to 
the emitter, thus providing a low resistance path between the ends of the 
transistor. Permanent damage to the transistor may result when punch 
through occurs. The punch through voltage may be appreciably less than 
the avalanche breakdown voltage experienced in a junction diode. 

While holes are being injected across the emitter junction into the base 
(assuming a p-n-p transistor), electrons are being injected from the base 
into the emitter. The total injection current across the emitter junction 
is the sum of these injection currents. Only the injected holes are effective 
in causing carriers to flow into the collector, however. The ratio of injec- 
tion current density from emitter to base to the total injection current 
density across the emitter junction is known as emitter efficiency rj. Then 

V = Jp (4.14) 

where J p is the current density resulting from injected holes and /„ is the 
current density resulting from injected electrons. In order to have rj 
approach unity, J P must be large in comparison with /„. 



Basic Amplifiers 123 

The diffusion current density which flows as a result of a positive charge 
concentration (Eq. 3.10) is 

J P =-qD v ^ (3.10) 

dx 

The diffusion constant D p has been evaluated for different types of 
material and the values for silicon and germanium at room temperature 
were listed in Chapter 3. Therefore, J v can be calculated if the rate of 
change of charge concentration with axial distance, dp/dx, can be found. 
Referring to Fig. 4.7, the initial injected charge density in the base is 
assumed to be/>„- The density at the beginning of the depletion region near 
the collector junction is essentially zero because the carriers are swept into 
the collector region at that point. In reference to Eq. 3.10, dp/dx must be 
constant throughout the base if the diffusion current is constant through 
that region. Actually, this would be the condition if there were no 
recombinations. With no recombinations, then, there must be a linear 
decline of charge concentration from p at the emitter junction to zero at 
the collector junction. This distribution is represented by the dashed line 
in Fig. 4Jb. The slope of this line is dp/dx. Then 

d -P . - So ( 4.1 5) 

dx w 

and 

/,-«*£? (4.16) 

w 

The actual variation of p with x, including recombinations, is the solid 
line of Fig. 4.7b. 

Let the injected electron density at the junction in the emitter region be 
n . If it is assumed that the emitter length is longer than the diffusion length 
L, the injected carrier density reduces to nje at distance L from the 
junction. 2 The slope of the density vs the distance curve at the emitter 
junction is 



(4.17) 



dn _ n 
dx L 

This situation is illustrated in Fig. 4.8. Then from Eq. 3.9 

J n = i^ (4.18) 

2 As the negative carriers recotnbine in the emitter region, the diffusion current due to 
these negative carriers also decreases. However, the charge which tends to accumulate 
as a result of the recombinations causes a drift current to flow in the emitter region. The 
sum of the drift current and diffusion current produces a continuous current through 
the emitter. 



124 



n a D n w 



Electronic Engineering 



(4.19) 



The ratio of current densities is 

Jp PoD P L 

In the interest of obtaining a high transport factor y, the base width 
w must be small in comparison with L. In addition, the number of injected 
carriers is proportional to the carrier density in the material from whence 
the injected carriers came. Therefore, n is small in comparison with p 
because of the light doping of the base region. Thus, the factors which 
cause the transport factor to approach unity also cause the emitter 
efficiency to approach unity. Therefore, the emitter efficiency also increases 
as the collector voltage is increased. 

Another factor which may affect a 
is called collector efficiency, <5. This 
factor accounts for the carrier multi- 
plication which may occur in the col- 
lector region. The collector efficiency 
is essentially unity except for rather 
high collector voltages which cause 
carrier multiplication by the process of 
ionization. This ionization results 
from high-velocity carriers as discus- 
sed in Chapter 3. The collector effi- 
ciency, d, will then exceed unity. In 
fact, it may quickly increase to infinity 
as avalanche breakdown occurs. A 
typical plot of <5 as a function of collector voltage is given in Fig. 4.9a. 
The collector efficiency may be expressed by the formula 

1 




0.37no 





Emitter 
junction 

Fig. 4.8. Injected electron concen 
tration as a function of x. 



where V A is the avalanche breakdown voltage and N is an empirically 
determined exponent (usually about 3 for an alloyed junction transistor). 
The current ratio a is the product of the emitter efficiency, transport factor, 
and collector efficiency. 

a = rjyd (4.21) 

Alpha (a) is known as the current amplification factor of a transistor in the 
common-base configuration. The variation of a with collector voltage is 
sketched in Fig. 4.9. The reduced values of a at very low values of collector 
voltage occur because minority carriers accumulate at the collector junc- 
tion. This accumulation occurs because the collector voltage is so low 



Basic Amplifiers 



125 




V c *■ 

<a) 

Fig. 4.9. (a) Collector efficiency as a function of collector voltage, (b) The variation of 
a with collector voltage. 

that the iR drop in the collector region causes a slight forward bias across 
the collector-base junction and carriers are injected from the collector to 
the base region. 

A scrutiny of this premise that the current flow through the base is due 
entirely to diffusion is in order. The charge density distribution (in the base 
of a typical transistor) which was given in Fig. 4.7b is enlarged and pre- 
sented as curve;? in Fig. 4.10. This injected charge tends to create a net 
positive charge and a resulting electric field in the base of&p-n-p transistor. 
This electric field causes free electrons in the base and connecting conduc- 
tors to flow in a direction that neutralizes the charge. If the charge were 
completely neutralized, the excess electron density distribution would be 
indicated by the dashed curve ri in Fig. 4.10. There would then be no 
net field in the base and current flow would be entirely by diffusion as 
assumed. Note that the excess electrons must be furnished by the external 




Fig. 4.10. Charge distribution in the base of&p-n-p transistor. 



12$ Electronic Engineering 

circuit because the base was electrically neutral before the charge injection 
occurred. This transfer of charge produces a capacitance effect which will 
be considered later with the high-frequency performance of the transistor. 
The free electrons tend to diffuse through the base in a manner similar to 
the diffusion of holes. The electrons cannot pass into the collector, 
however, because the collector barrier forces them back into the base. 
Consequently, there is a negative charge density at the collector junction. 
In addition, the electron density is not sufficient to neutralize the positive 
charge near the emitter junction because of the electron diffusion. The 
actual negative charge distribution then follows the solid curve labeled n 



Fig. 4,11. Variation of a with emitter current. 

in Fig. 4.10. The net charge distribution in the base is the difference 
between/? and n as shown by the shaded area in Fig. 4. 10. This unneutral- 
ized charge creates an electric field which opposes the electron diffusion. 
The field builds up to a magnitude which produces an electron drift 
current equal in magnitude and opposite in direction to the electron 
diffusion current. 

The electric field in the base region is in a direction that accelerates the 
holes through the base. In other words, the hole diffusion is aided by 
drift due to the electric field. The transport factor is therefore increased 
by the net charge distribution in the base, and the charge distribution is in 
turn a function of the emitter current, because the injected charge density 
is proportional to the emitter current. The transport factor therefore 
increases with emitter current. 

The emitter efficiency decreases as the concentration of electrons near 
the emitter junction increases because the rate of injection of electrons into 
the emitter is proportional to the free electron density on the base side 
of the junction. As the emitter current is increased, the transport factor 
increases more rapidly than the emitter efficiency decreases for small values 
of emitter current, but as the emitter current becomes larger, the emitter 
efficiency decreases more rapidly than the transport factor increases. 
Thus a is maximum at some specific value of emitter current in any given 



Basic Amplifiers 



127 



transistor. A sketch of a as a function of emitter current is provided in 
Fig. 4.11 in which the variation of a is exaggerated as compared with a 
typical transistor. 



4.5 EQUIVALENT CIRCUITS 

As with the diodes, equivalent circuit's approximating the behavior of 
amplifiers have been developed. The basic ideas for these equivalent cir- 
cuits may be derived from Fig. 4.12. The amplifier usually contains two 
input terminals (or an input port) and two output terminals (or an output 
port). (One input terminal and one output terminal may be common.) 



h — >- 



t 

Vi 



-ol 



2o- 



Electrical 
network or 
amplifier 



-ol 



2o- 



-• — h 



\ 



Fig. 4.12. A general four-terminal network or amplifier. 

A circuit of this type is usually known as a two-terminal-pair network (or 
as a two-port network). Regardless of conditions inside the network, 
only four quantities can be measured in the external circuits. These quan- 
tities are input voltage and current (V t and 7 X ) and output voltage and 
current (K 2 and 7 2 ). 

If the currents I x and 7 2 are assumed to be the independent variables, the 
voltage F x may be a function of both 7 X and 7 2 . Thus, if terminals 2-2 are 
open circuited (7 2 = 0) and a Current 7 X is applied to the input terminals 
1-1, a voltage K x (and also V 2 ) can be measured. The ratio V x \l x can be 
expressed as an impedance Z n . In a simple d-c situation, Z u may be a 
single resistance. In contrast, if V x and I x are expressed in the Laplace 
transform form, the impedance Z u may be a function of s. (For sinusoidal 
a-c signals, Z u will be a function of joy.) Now, the input terminals 1-1 
are open circuited and a current 7 2 is applied to the output terminals 2-2. 
Again, the voltage V x (and also V^ is measured. The ratio of VJI % is 
expressed as an impedance 3 Z 12 . Thus, we may write 

Vi = ZiJi (4.22a) 

and 

Vj. = Z 12 7 2 (4.22b) 

• Since Z„ involves the current and voltage at the same set of terminals, this imped- 
ance is called a self impedance. In contrast, Zu involves the current at one set of termi- 
nals and the voltage at a second set of terminals and is known as a transfer impedance. 



128 



Electronic Engineering 



when both I t and I 2 are present and Z n and Z 12 are linear the law of 
superposition permits us to combine the foregoing equations and express 
F^as 

V, = Z xx h + Z^h (4.23) 

Similarly, the ratio of V^h with the output open can also be found. 
This ratio is called Z 21 . Also, the ratio of VJI 2 with the input open is 



h 


Zi 




7„ 


h 


> t 






\ 






Vi 


Z\i 


v 2 


i> 






o 



(a) 



J-T- * 




-rv 


- z 2 


^ 




hZ m 


^ .1 

h 


1* 




^i 


Zl2 


v% 









— — — — o 



Fig. 4.13. (a) An equivalent T circuit for Fig. 4.12 where the network contains only 
R, L, and C components. (6) An equivalent T circuit for Fig. 4.12 where the network 
contains an amplifier. 

found and is called Z 22 . When these ratios are established, we can write 

V 2 = Z 2 J X (4.24a) 

V 2 = Z 22 I 2 (4.24b) 

These equations, if linear, are combined (by the use of the law of super- 

V 2 = Z 21 /! + Z 22 / 2 (4.25) 



and 



position) to yield V 2 . 



If the circuit contains only R, L, and C components, the circuit is known 
as a passive circuit and Z 12 will be equal to Z 21 . One equivalent circuit 
which represents Eq. 4.23 and Eq. 4.25 is given in Fig. 4.13a. The validity 
of this equivalent circuit may be checked by writing the loop equations. 



V x - (Z x + Z 12 )I X + Z 12 I 2 
V 2 ^Z 12 I 1 + (Z 12 + Z 2 )I 2 



(4.26) 
(4.27) 



Basic Amplifiers 129 

These equations are the same as Eqs. 4.23 and 4.25 if 

Z 12 = Z 21 , (Z a + Z 12 ) = Z u and (Z 12 + Z 2 ) = Z 22 

This configuration is known as the f configuration for obvious reasons. 
In an amplifier circuit, Z 12 is not equal to Z 21 . This circuit is known as an 
active circuit. Either a voltage or current generating source can be added 
to the passive equivalent circuit to account for the difference between Z 12 
and Z 21 . The circuit of Fig. 4.13a, must, therefore, be modified when the 
circuit is active. A modification is shown in Fig. 4.136 which correctly 
represents an amplifier. The loop equations for the circuit of Fig. 4. 1 2>b are 

Vi = ( z i + z i2)A + Zuh (4.28a) 

V % = (Z 12 + ZJ/j + (Z 2 + Z 12 )/ 2 (4.28b) 

The term Z m has been added to Z 12 to form Z 21 . The term Z m may be 
either positive or negative to fit a given situation. The voltage generator 
I t Z m also indicates how amplification may be accomplished. 

As already noted, the values of the Z parameters can be found as 
follows : 

Z n = the input impedance when the output circuit is open. 

Z 12 = the ratio of the open circuit voltage at the input terminals to the 

current flowing in the output circuit. 
Z 22 = the output impedance when the input circuit is open. 
Z 21 = the ratio of the open circuit output voltage to the input current. 

Some limitations on the accuracy of the equivalent circuits do exist. 
The impedances are assumed to be linear impedances and, as the reader 
has no doubt already observed, the characteristics of the tubes and 
transistors are not linear. However, for small-signal variations, the 
characteristics are essentially linear. In fact, the smaller the signal, the 
more nearly linear are the characteristics of the devices over the range of 
operation. Consequently, the equivalent circuits which have been developed 
and those yet to be developed in this chapter are valid for small-signal 
amplifiers. Larger signals can be handled if the characteristics of the 
device are approximately linear over the region of operation. 

An equivalent T circuit for a transistor may now be drawn. Figure 4.14 
shows the grounded base configuration and an equivalent circuit. The 
impedances which are essentially pure resistances at low frequencies are 
given subscripts corresponding to the terminals to which they connect. 

Thus r e is essentially the dynamic resistance of the forward biased 
emitter junction plus the small ohmic resistance of the fairly heavily doped 
emitter region. The dynamic junction resistance can be easily determined 



130 Electronic Engineering 

as a function of the bias value of the emitter current. The voltage-current 
relationship across the emitter-bias junction is expressed by the diode 
equation (Eq. 3.24) 

'* = ~ W* 9(W * T - 1) (4.29) 

The dynamic conductance g t = l/r e = di E jdv EB evaluated at the bias 



r 




(a) 



E 

-o— 



-wvw- 



y c 



-o — ^J-^-WWv* — o—o-o 



TmU 



m 

Fig. 4.14. An equivalent T circuit of a grounded base transistor. 

value of emitter current I E . Then, taking the derivative of Eq. 4.29 with 
respect to v EB , 

ge=~ -^iEoe™**" 1 * (4.30) 

But, using Eq. 4.29, 

-I EO e m »* /kT = i E - I EO (4.31) 

When the emitter junction is forward biased 0. 1 volt or higher, i E » I eo . 
Then 



1 ■ Q T 

Se kT E kT E 



(4.32) 



(4.33) 



where I E is the Q point, or bias value of i E . Thus 

kT 

r ~ 

qIe 
At 17°C, qJkT =40,g e = 40I E (amp) and r e = 25// E (ma). 

The resistance r b accounts for the effect of collector voltage variations 
on emitter current as well as the ohmic resistance of the lightly doped 



Basic Amplifiers 131 

base. The collector voltage influences the emitter current because of the 
variation of effective base width with varying collector voltage. For this 
reason, r b is frequently known as the base spreading resistance. Finally, 
the resistance r e includes the effect which the varying collector voltage, 
and hence varying base width, has on the collector current, as well as the 
surface leakage resistance across the reverse biased diode. Also, r e includes 
the effect of current multiplication due to avalanche effect in the collector- 
base depletion region. When high frequencies are considered, the equiv- 
alent circuit must also include capacitances. These will be treated in detail 
in Chapter 7. At low frequencies, the loop equations of the two loops of 
Fig. 4.14 are 

v t = (r. + r b % + r b i c (4.34a) 

v. = (r m + r b )i e + (r b + r e )i c (4.34b) 
When these equations are compared to Eqs. 4.23 and 4.25, 

Zu = r e + r b (4.35a) 

Z 12 = r b (4.35b) 

Z ta = r b + r m (4.35c) 

Z 22 = r c + r b (4.35d) 
Combining Eqs. 4.35b and 4.35c, 

r m =Z 21 -Z 12 (4-36) 

A useful variation of the equivalent circuit shown in Fig. 4.146 is given 
in Fig. 4.15a. In this variatipn, the voltage generator between the points 
x and y (Fig. 4.146) is replaced by an equivalent current generator. The 
parameters of this current generator are found by applying Norton's 
Theorem to the generator between points x and y of Fig. 4.146. When 
the output circuit of Fig. 4.15a is shorted as shown in Fig. 4.156, the 
direction of collector current will be as shown. Since the short circuit 
collector current will be a times as large as i t , Kirchhoff's current law yields 

<*■% + *. = ~U = *h ( 4 - 37 > 

Kirchhoff's voltage law applied to Fig. 4.156 gives 

//-. - hr b = (4.38) 



so 



i. = ^ (4.39) 



132 



n? 



-WWvV 



n 



■WWW 1 



■& 



Electronic Engineering 

ait 



<■>■— VWW - ^ 



. n 



(a) 



■e 



Otl e 




■vww- 



(b) 
Fig. 4.15. A current source equivalent T circuit. 



!"■ 



But, i b = (1 — a)i e since only small signal variations are being considered. 
Therefore, 



K = 



(1 — a)i e r b 



(4.40) 



When this value of i x is substituted into Eq. 4.37, 

,, , (1 - a)t> 6 



or 



a' = a - (1 - a) ■ 



(4.41) 



(4.42) 



In the usual transistor, r b « r c and (1 — a) « 1 . Therefore, a' is practically 
equal to a. 

The relationship between the voltage generator of Fig. 4.146 and the 
current generator of Fig. 4. 1 Sa can now be found. If these two generators 
are equal, the open circuit voltages and the short circuit currents must be 
equal. Then, 

r mh = «Vc (4.43) 

or 

r m = «> c ^ o^c (4.44) 



Basic Amplifiers 133 

Table 4.1 lists a set of parameters for a typical low-power transistor. 

TABLE 4.1 
Parameters for a Typical Low-Power Transistor 



a = 0.984 
r/=13Q 
r„ = 1000 a 
r c = 1.8 megohm 



PROB. 4.6. Find the value of a' for the transistor of Table 4.1. 
near the value of a? Answer: a' = 0.9839912. 



Is this value 



Vi 



I2Z12 





h 



hZn 



Vi 



Fig. 4.16. An alternate z-parameter equivalent circuit. 



Another equivalent circuit known as a z-parameter circuit which may 
be derived from Eqs. 4.23 and 4.25 is shown in Fig. 4.16. A check of 
the loop equations verifies that this is a valid equivalent circuit. 

Actually, as the reader may have surmised, a large number of equivalent 
circuits can be devised. For example, if V Y and K 2 of Fig. 4.12 are assumed 
to be the independent variables, a y-parameter circuit can be developed. 
Thus, if terminals 2-2 are shorted and a voltage V x is applied to terminals 
1-1, the relationships I l jV 1 = Y n and I i jV l = Y 21 can be measured. 
Then, if terminals 1-1 are short circuited and a voltage V 2 is applied to 
terminals 2-2, the relationships IJ V 2 = Y 12 and I J V 2 = Y 22 can be 
determined. Then, by the law of superposition we can write the two 
equations 

h = rii^i + ^12^2 (4.45a) 



h= ^1+ ^22^2 



(4.45b) 



One form of the y-parameter equivalent circuit is given in Fig. 4.17a. 
Nodal equations can be written to verify the validity of this equivalent 



134 



Electronic Engineering 



circuit. Again, we may summarize the method of measuring the 
y-parameters. 

F u = the input admittance with the output terminals shorted. 

Y 1Z = the ratio of the short circuit current through the input terminals to 

a voltage which is applied to the output terminals. 
Y 21 = the ratio of the short circuit current through the output terminals 

to a voltage which is applied to the input terminals. 
Y 22 = the output admittance when the input circuit is short circuited. 




h- 



Y12V2 



d) 



'.Y22 v 2 



Vl 



Y21W 



1 MA- 

G22 



IG12V2T- 



-h 



v 2 



(a) (b) 

Fig. 4.17. (a) The equivalent Y circuit, (b) The equivalent G circuit. 

An additional circuit can be found by assuming I 2 and V x are the 
independent variables. The two resultant equations are 

h = G 11 V 1 + G 12 h (4.46a) 

V 2 = G 2X V^ + G 22 I 2 (4.46b) 

One equivalent circuit which represents Eq. 4.46 is shown in Fig. 4.17b. 
The ^-parameters are denned as follows : 

G lt = the input admittance with the output terminals open. 

G 12 = the ratio of the short circuit input current to the current which 
is applied to the output terminals. 

G n = the ratio of the open circuit voltage at the output terminals to the 
voltage applied to the input terminals. 

G 22 = the output impedance when the input is short circuited. 

The final equivalent circuit to be discussed here is known as the hybrid 
or A-parameter circuit. This circuit receives its name from the fact that it 
is a hybrid combination of the z- and y-parameter circuits. The /(-param- 
eters of transistors are most easily measured and therefore are given by 
most transistor manufacturers (^-parameters are given quite often for 
high-frequency applications). 

In developing the A-parameter circuit, the independent variables are 
assumed to be / x and V 2 . Then, the two characteristic equations have the 
form 



Vi = #llA + #12^2 

I 2 = H 2 ili + H 22 V 2 



(4.47a) 
(4.47b) 



Basic Amplifiers 



135 



where H Y1 = the input impedance when the output is shorted. 

H l2 = the ratio of voltage appearing at the open circuit input 

terminals to the applied voltage at the output terminals. 
H 21 = the ratio of current appearing at the short circuited output 

terminals to the input current. 
H 22 = the output admittance when the input is open. 
Previous equations have used capitals to indicate voltages, currents, 
impedances, and admittances. This notation follows the IEEE recom- 
mendations for d-c or steady-state a-c signals (see pages xx and xxi of the 



H 1 






■< ^ 


R = h i '< 








n T 




) A /*i < 


>G = h „ o 


o ' 


1 




o 



Fig. 4.18. An equivalent A-parameter circuit. 



List of Symbols). When time-varying signals are considered, lower case 
letters are used. In addition, the h parameters as defined by the IEEE 
carry letter subscripts (capital subscripts for d-c values and lower case 
subscripts for time-varying signals). Thus, the preceding characteristic 
equations (Eqs. 4.47a and 4.47b) when applied to a small-signal variation 
about a fixed operating point in a transistor have the form : 



»i = hji + h r v 
h = V« + h » V o 



(4.48a) 
(4.48b) 



where h t = the input impedance when the output voltage v is held con- 
stant and thus v„ = 0. 
h = the output admittance when the input current is constant. 
h r = the ratio of voltage induced in the input to the output voltage. 
It is called the open-circuit reverse voltage transfer ratio. 
h, = the ratio of short circuit output current to the input 
current. This is called the forward current amplification factor. 
When applied to the common-base transistor, this parameter 
is —a. 
An equivalent circuit represented by these equations is shown in Fig. 
4.18. 



136 



Electronic Engineering 



One set of parameters may be obtained from another set of parameters 
by matrix manipulations. To aid the reader in making these manipulations, 
Table 4.2 is included. 

TABLE 4.2 

Formulas for Converting Parameters 



Con- 
versions 
from 




A represents the term (Az = z xx z 2i - 


— Z 12 Z 2 i) 




z's 


z 22 

3/ii = -t- 

Az 

— z 21 
2/21 = -7— 

Az 


— z la 

Zll 

2/22 ~ Az 


fill = 

Z 2 2 

fill = 

2 22 


Al2 = 

Z 2 2 

A 22 = 

z 22 


1 

#11 = — 

Zll 

«J1 

^11 = — 

Zll 


z 14 

8» = — 

z ll 
Az 

#22 = — ■ 

z ll 


y's 


2/22 

Zii = -r- 

At/ 

-2/21 

221 Ay 


-2/12 

212 Ay 

2/n 
222 "A^ 


Ay 

2/22 
-2/21 
g%1 ~ 2/22 


2/12 

#13 = rr 

2/22 

1 

<?"= — 
2/22 


*ll=-i 

2/ii 

■ , 2/21 
«21 = 

2/n 


, -2/i2 

"12 = 

2/n 
* - Ay 

«22 = 

2/n 




fin 

—ftn 
^ 1 = -aT 


— A l2 
^ 12 ~ AA 

An 


AA 

Zll = — 
#22 

— A 21 


Al 2 

Z 12 — 7~ 

"22 

1 

A 2 2 


1 

2/11 = T- 

An 

2/21 = 7— 

All 


~Al2 


h's 


2/l2 , 

AA 

2/22 _ ATi 




221 /. 

«22 


#'s 


"ll = T— 

A? 

"21 1 

A? 


A - _lfl2 

"12 . 

A? 

h - gl1 
«22 — T- 

A? 


A? 

2/n = — 

#22 

2/ 2 i - _ ^ 21 

#22 


2/l2 

#22 

1 

2/22 = 

#2 2 


1 

Zll = — 

#11 

Z 2 1 

#11 


z - ~ gl * 

z 12 — 

#11 

z -^ 

z 22 — 

#n 



PROB. 4.7. The following measurements are taken on a two-port device: 

Test 1 

The output terminals are shorted and the following quantities are measured: 
V-l = 0.5 v, / x = 1 ma, I 2 = 50 ma. 

Test 2 

The input terminals are short circuited and the following quantities are 
measured: I x = 100 fi amp, / 2 = 10 n amp, V 2 = 10 v. Develop an equivalent 
circuit which could be used to represent this device. Determine the magnitude 
of the parameters in this equivalent circuit. 

PROB. 4.8. The following tests are conducted on a two-port device : 

Test 1 

The input terminals (1-1) are open circuited, V 2 = 10 v, V x = 0.01 v and 
/ a = 10 fi amp. 



Basic Amplifiers 137 

Test 2 

The output terminals (2-2) are open circuited, V 1 = 0.1 v, I x = 1 mamp, 
V 2 = 10 v. 

(a) Draw an equivalent circuit for this device and indicate the magnitude of 
all parameters in the circuit. Answer: Z n = 100 il, Z 12 = 1 KQ, Z 21 = 10 Kil, 
Z 22 = \MCl. 

(Z>) Convert this equivalent circuit to an /(-parameter equivalent circuit. List 
the values of all parameters. 

The h parameters for a transistor may be obtained from the characteristic 
curves of Fig. 4.19. (Remember that the equivalent circuits are valid for 
small-signal variations only.) The parameter h may be denned for the 
common base configuration 4 as 

di c 



hot = , 



(4.49) 

if—constant 



The value of h 00 at a given operating point is, therefore, equal to the slope 
of the characteristic curve at the given point on the set of collector charac- 
teristics. This slope is represented by the line AB of Fig. 4.19. The reader 
may question the value of 3 ma emitter current when the input circuit is 
supposed to be open. Remember, however, that the equivalent circuit is 
valid for the small signal component of current. Hence, if the time- 
varying component of emitter current is zero (emitter current is constant), 
the emitter circuit is effectively "open circuited" for the small signal 
component. 

The parameter A/a) can be defined for the common base configuration 
as 

di c 



h fb = ■ ,. 
di f 



(4.50) 

V£=constant 



This equation represents the slope of a curve which could be drawn by 
plotting i c as a function of i E with the collector voltage held constant. The 
required currents and voltages are also given in the collector characteristics 
of the transistor. The line CD of Fig. 4. 19 represents a constant collector 
voltage v c . The ratio of the length of this line as measured by i c to the 

4 Equation 4.49 defines the output admittance as the ratio of the change of collector 
current to the change of collector voltage with the emitter current maintained constant. 
This definition is obviously correct for small-signal variations. 

A double subscript notation is used for transistor parameters. The first subscript 
indicates which parameter is involved and the second subscript indicates which element 
of the transistor is used as the common element to both input and output circuits. Hence 
h„ b indicates the output admittance, with the input current constant, for a common-base 
configuration. 



138 



Electronic Engineering 



-6 



-5 



-4 



£ 

J= -3 

1 



-2 



-1 



1 














*• 










6 ma 














_^^^_^^_ 












5 ma 
















( 




.D 






4 ma 




r 


A 


B 
C 






3 ma 




( 










2 ma 




[ 










jj5= 1 ma 



+1 



0.6 



0.5 



0.4 



.£ 0.3 



0.2 



0.1 



-5 



-10 -15 -20 -25 -30 

v CB in volts 





















h^^ZE- 








v CB =0v 


si, 


-$^^ ( 










-lOvyi 


V-20v 









































12 3 4 5 6 

i E in milliamperes 

Fig. 4.19. Characteristic curves of a common-base transistor. 



Basic Amplifiers 



139 



length of this line as measured by i E gives approximately 5 the value of h n 
in this region. 

Unfortunately, it is difficult to measure the currents accurately enough 
to determine an accurate value of h n (or h oh for that matter) from these 
curves. A technique for determining h n (and h ob ) with greater accuracy will 
be given in Chapter 5. Note also that h fb is negative because i E is positive 
and i c is negative. 

The parameter h t can be written for the common-base configuration as 



h ih = 



dv E 
di E 



(4.51) 



i>£=constant 



Each curve of the input characteristics was plotted for a particular, constant 

collector voltage. Therefore, the slope of these curves represents h ih . 

The line ^(Fig. 4.19) represents the value at the chosen operating point. 

The parameter h r can be written for the common-base configuration as 



K h = 



dv E 

dv n 



(4.52) 



iE— constant 



This parameter is the slope of the curve obtained by plotting v E as a 
function of v c with the emitter current held constant. A vertical line on 
the input characteristic curves represents a constant emitter current. The 
line GH of Fig. 4. 19 will yield an approximate value for h ri at the indicated 
operating point. The length of this line in terms of emitter voltage divided 
by the length of this line in terms of collector voltage gives the approximate 
value of h rb . 

When Eqs. 4.49, 4.50, 4.51, and 4.52 are considered, two quantities 
affect v E . These quantities are i K (Eq. 4.45) and v c (Eq. 4.52). Also, two 
quantities affect i c . Again these quantities are i E (Eq. 4.50) and v c 
(Eq. 4.49). Then, by superposition the voltage and current relationships 
in the transistor may be expressed as 



a dv E 

di E 

di c 
di E 



A/ K + 



»e-fc 



Air 



Ai E + 



dv E 
dv c 

di c 

dv r 



Av r 



Atv 



(4.53) 



(4.54) 



i£=k 



v c =k " V C 

But, for small-signal variations, the small change of signal Av E can be 
equated to the time-varying component of voltage v e . The input voltage v { 
in Fig. 4.18 is equal to v e in the common-base configuration. Similarly, 



5 In this approximation, the ratio — - 



is taken rather than — 

■ c = constant "'.£ 



v c = constant 



140 Electronic Engineering 

we note that A/ E = i it Av c = v , and Ai c = i . When these values are 
substituted into Eq. 4.53 and Eq. 4.54 and the h parameters are substituted 
for the derivative terms, these equations become 

v, = Kbh + KiPo (4.55) 

h = n nh + K»Vo (4.56) 

These equations are the same as Eq. 4.48. Hence the hybrid equivalent 
circuit is valid for the transistor. 

The foregoing treatment indicates that the hybrid equivalent circuit 
parameters can be found from the collector characteristic and input 
characteristic curves. 

PROB. 4.9. Obtain the common-base h parameters for the transistor of Fig. 
4.19 at the operating point indicated. Note the difficulty of obtaining accurate 
values for h ob and h fb . Answer: h ob ~ 2 x 10~ 6 mho, h ib ~ 50Q, h rb ~ 3 x J0~ 3 , 
h fb ~ -0.98. 

PROB. 4.10. Using the h parameters of Prob. 4.9, calculate the voltage and 
power gains of a common-base circuit which has a load resistance of 10 KO. 

4.6 INPUT AND OUTPUT IMPEDANCE OF THE 
COMMON-BASE CONFIGURATION 

The equivalent circuits will be used to illustrate the method of solution 
for transistor circuits. In order to calculate the input impedance of a 
transistor with the output shorted, the circuit of Fig. 4.20a is drawn. A 
voltage v t is applied to the input and the input current i e is calculated. 

v t = i e r e + 0e + Qr b (4.57) 

but with the output shorted, 

i c = — *»« 
and 

»i = hr e + (1 - a)v'« (4.58) 

Then, since h ib is the input impedance with the output shorted, 

*» = - = r. + (1 - aK (4.59) 

Thus, h ib is obtained in terms of the equivalent T parameters. From 
the equivalent circuit, it may be observed that the input impedance with 
the output open is r e + r b = Z u . These values are the minimum and 
maximum values of input impedance for a given common-base transistor 
at the specified operating point. Intermediate values are obtained for 
finite values of load resistance. 



Basic Amplifiers 



141 



The h parameters h ob and h rb may be obtained in terms of the r param- 
eters with the aid of Fig. 4.206. In this figure, the input is open and 
voltage v c is applied to the collector. Since /„ = 0, the output resistance is 
Z 22 = r c + r b and 

1 1 1 



^22 r c T r b 



(4.60) 



The voltage v c applied to the output terminals is i c (r c + r b ) and the 



VvVHO^- 




-WW 



^_^ i c 
-AAA/ — zO+— <? 



n 



(b) 

Fig. 4.20. Equivalent circuits for determining the relationship between r and h 
parameters. 

voltage across the open circuit input terminals is i c r b . Therefore 

'c r b r b 



h rb = 



— "ob r b 



(4.61) 



i c (r e + r b ) r c + r b 

As previously mentioned, h fh = —a. and from Eq. 4.44, a ~ r m /r c . 

The input resistance, output resistance, voltage gain (r„/r,), and current 
gain (iJQ of a transistor can be determined for any value of load resis- 
tance or source resistance by the solution of any valid equivalent circuit. 
The /r-parameter circuit will be most commonly used in this text. For 
example, the input resistance can be determined from the equivalent 
circuit of Fig. 4.21a (or Eq. 4.48A) by noting that —i c — vjR L . Then, 



-h fb i e = v c lh ob + — j 



(4.62) 



The loop equation for the input circuit is 

f i = 'Aft + h rbVc ( 4 - 63 ) 

When Eq. 4.62 is solved for r c , and this value of t' c is substituted into Eq. 
4.63, Eq. 4.63 becomes 

v t = ij h ib — \ (4.64) 



142 

Then, 



Electronic Engineering 

(4.65) 



7 _£?_/. K„h fb R L \ 

i e \ 1 + h ob R L / 

Although Eq. 4.65 is very useful, the reader should develop skill in solving 
the circuit equations rather than attempting to memorize the resulting 
equations. 

By the same type of reasoning, the output impedance of a transistor 
amplifier can be found. In this case, the input signal voltage v t is reduced 




Fig. 4.21. Equivalent h circuit for determining input and output impedance. 

to zero but the generator resistance R g is still present across the input 
terminals. Then, an equivalent circuit is drawn as in Fig. 4.216. In this 
circuit, the nodal equation for terminal C is 

h = VcKi + h fh i e (4.66) 

Also, the loop equation for the emitter loop is 

KM = -ie(h ib + R g ) (4.67) 

When Eq. 4.67 is solved for i e and this value is substituted into Eq. 4.66, 



Z„ = - c = 



1 



h ib + R g 



U (u ft/Ah \ 
\ Kb - h ib + rJ 



h„bhih + h ob R„ — h fb h r 



(4.68) 



Note that both h tb and /, are negative. Also observe that the output 
impedance approaches \jh ob as R a approaches an infinite value. However, 



Basic Amplifiers 143 

as R„ approaches zero, the output impedance becomes much smaller than 
l/Zi^. In general, however, the input impedance is low and the output 
impedance is high for a common-base transistor. For emphasis, this low 
input impedance is due to the emitter-base junction being biased in the 
forward direction. Similarly, the high output impedance is due to the 
collector-base junction being biased in the reverse direction. 

PROB. 4.11. From the equivalent T circuit, calculate the input impedance of the 
transistor of Table 4.1, when the load resistance is (a)0 CI, (b) 10 KQ. Answer: 

(a) 28 CI, (b) 29 CI. 

PROB. 4.12. From the equivalent T circuit, calculate the output impedance of 
the transistor of Table 4.1, when the driving source resistance is (a) 100 CI and 

(b) 100 Kii. 

PROB. 4.13. Derive an expression for the voltage gain of a common-base 
transistor amplifier as a function of load resistance using the A-parameter 
equivalent circuit. (Hint: Use Eqs. 4.62 and 4.63.) Answer: G v = —*/»/(*<*/!» 
— h rb h fb + h ib Gi). 

PROB. 4.14. Derive an expression for the current gain of a common-bast; 
transistor amplifier as a function of load resistance using the A-parameter 
equivalent circuit (Fig. 4.21a). 

4.7 THE TRIODE TUBE 

The triode tube was invented in 1906 by Dr. Lee DeForest, who inserted 
a third element in the form of a wire mesh or screen between the cathode 
and plate of a high vacuum diode. The configuration which resulted is 
shown in Fig. 4.22. In this device, the electrostatic field between the 
control grid and the cathode is able to control the flow of electrons to the 
plate. 

The action of the control grid can be visualized with the help of Fig. 4.23. 
Electrons are emitted by the hot cathode. Since the plate is maintained 
positive with respect to the cathode, an electrostatic field exists between 
the plate and cathode. The potential on the control grid "warps" this 
electrostatic field. If the control grid is positive, (Fig. 4.23a), most of the 
electrons near the cathode are attracted toward the plate. Since the 
control grid is a wire mesh, most of the electrons pass through the control 
grid to the plate. However, a few of the electrons are intercepted by the 
positive grid wires. The current through the tube then has the relationship 

i K = h + ip (4.69) 

where i K is the cathode current, i a is the grid current, and i P is the plate 
current. 



144 




Electronic Engineering 

Control grid 



ft 

Fig. 4.22. A triode tube. 

If the voltage on the control grid is at the same potential as the cathode 
(Fig. 4.23b), a fewer number of electrons near the cathode are attracted 
toward the plate. The reason for this reduced current (as compared to the 
positive grid case) is due to the reduced electric field intensity near the 
cathode. The electric field intensity is equal to —dVjdx where x is distance. 
Therefore, the slope of the curves in Fig. 4.23 is a measure of the electric 
field intensity. A reduced electric field intensity results in a reduced flow 
of electrons, as discussed in Chapter 1. When the control grid is at the 
same potential as the cathode, the number of electrons collected by the grid 
is very small. Therefore, i a of Eq. 4.69 is almost equal to zero for a grid 
potential of zero volts above the cathode. 




Fig. 4.23. A plot of the potential distribution inside a triode tube, (a) The control grid 
positive with respect to cathode; (b) the control grid at the same potential as the cathode ; 
and (c) the control grid more negative than the cathode. 



Basic Amplifiers 



145 



When the potential of the control grid is negative with respect to the 
cathode (Fig. 4.23c), the plate current is reduced even further. Also, since 
the control grid is negative, the electrons are repelled by the control grid 
and the grid current i a is essentially equal to zero. However, plate current 
may still pass between the grid wires. 

As indicated previously, the control grid is able to control the amount of 
current to the plate. In fact, the action is very similar to the action in a 
transistor. The electric field between the cathode and control grid region 
is analogous to the electric field in a transistor between the emitter and 
the base. Also, the electric field between the plate and control grid of a 
triode is analogous to the electric field 
between the base and collector of a tran- 
sistor. Essentially no positive carriers exist 
in a high vacuum tube, so only electrons 
act as carriers. The triode does have an 
advantage over the transistor inasmuch as 
the control grid draws essentially no current 
if maintained more negative than the cath- 
ode. 6 In this condition, the parameter cor- 
responding to « in the transistor is equal 
to unity. In addition, the control grid con- 
sumes essentially no power if it draws 
essentially no current. Consequently, the 

triode tube is usually operated with a negative control grid. Cases where 
the control grid does draw current will be investigated in Chapter 10. 

The symbol of a triode tube is given in Fig. 4.24. The similarity between 
this symbol and the actual tube is apparent. 

If the control grid current is negligible, only three variable input and 
output quantities remain. These quantities are: voltage between the 
cathodeand the control grid, plate (or cathode) current, and the voltage 
between the plate and the common electrode. The most commonly used 
triode circuit is the common cathode circuit to be investigated in Chapter 5. 
Consequently, practically all triode characteristic curves list the voltage 
between the plate and the cathode rather than the voltage between the 
plate and the control grid as a parameter. 

The characteristic curves of a triode tube may be plotted by using the 
circuit in Fig. 4.25. In this circuit the three meters indicate the required 
three variables. A set of curves which corresponds with the set of collector 
curves of a transistor is plotted. The plate current is plotted as a function 




Plate 
Envelope 
Control grid 

Cathode 
Filament 



Fig. 4.24. Symbol for a triode 
tube. 



6 As will be discussed later, some current does flow in the control grid circuit due to 
capacitance between the control grid and other tube elements. 



146 



Electronic Engineering 




Fig. 4.25. Circuit used for plotting triode characteristic curves. 



ID 






o/ 






























III 






























yt.l 


?J- 


















































































































"O/ 


1 


ij 


























i 




















12 




o/ 




























/ , 






10/ 
























/ / 






If 
























r / 
























O 




1 


/ 






1 


*>l 
















Q. 




1 


/ 








i L 
















£ 






/ 
























<o 




J 


/ 










c ">/ 




















/ 










~^i 














E 




1 


y 










ij 














.£ 8 




f 
















-7 








































.^ 




i 
















1 1 












c 






i 














1 












<L> 






i 














1 


V 


• I 








t 




I 


1 














1 


*■*. 










3 




1 


i_ 














/ 


/ 












































1 
















( 




•Sf/ 
































1 1 








Q_ 


























*/ 




4 


























1 






/ 






fc 






















/, 




/ 
























/ A 


'^- 


n 


A 























100 



400 



500 



200 300 

Plate voltage in volts 

Fig. 4.26. Characteristic curves for a 6.15 triode. (Courtesy of Radio Corporation of 
America.) 



Basic Amplifiers 



147 



of plate voltage. Since one curve is obtained for each different value of 
grid-to-cathode voltage, a family of these curves, which is known as a set of 
plate characteristics results. Figure 4.26 is a typical set of plate charac- 
teristics for a triode. The triode tube does not normally have a set of input 
characteristics since there are only three variables and all necessary 
information is obtainable from the collector or plate characteristics 
provided that v a is negative. 



4.8 THE TRIODE EQUIVALENT CIRCUIT 

The vacuum triode circuit which corresponds to the common-base 
configuration of the transistor is shown in Fig. 4.27. In the vacuum tube, 
the circuit is known as a grounded grid amplifier. The equivalent circuit 7 



+ 6 




Fig. 4.27. The grounded grid amplifier. 

for the triode is found by observing that the plate current i P is dependent 
on the control grid-to-cathode potential v GK , and also on the plate-to- 
cathode potential v PK . 

When the current is the dependent variable and two voltages are the 
independent variables, the situation is described by the y-parameter 
equations (Eq. 4.45). This form of equation can be used for i P but the 
only independent equation we can write for i K is i K = —i P . Thus, in 



7 A word on symbol nomenclature is in order at this point. In the early days of radio, 
batteries were used as the power sources. The A battery heated the filaments and the B 
battery supplied plate voltage. The C battery furnished negative control grid bias poten- 
tial. Accordingly, most literature and, in fact, the old IRE standards use the symbols i b 
and i e for the instantaneous total currents in the plate and control grid circuits of a 
vacuum tube. The instantaneous time-varying components of current used are i, and /,. 
In this text, to avoid confusion between base and collector currents and to conform to 
recent IEEE standards, /, and /„ will be used to indicate the total instantaneous currents. 
Similar subscripts will be used for voltages. 



148 



Electronic Engineering ■ 



general form, the characteristic equations are 

i K = — i P 
and 



(4.70a) 



ip = Y 2l v UK + Y 22 v PK (4.70b) 

In terms of time-varying components, these equations can be written as 

'* = -ip (4.71a) 



h = gm v ak + 



I 



"pk 



(4.71b) 



One form of equivalent circuit which represents Eq. 4.71 is shown in 
Fig. 4.28. Note that this circuit is similar to half the circuit in Fig. 4.17a, 



i-A/WvV-! 




Fig. 4.28. An equivalent circuit for the tube of Fig. 4.27. 

if the cathode is assumed to be the common element. Since both of the 
voltages in Eq. 4.71b use the cathode as reference, this result is not 
unexpected. 

The parameter g m (in Eq. 4.7 1 b) is known as the transconductance of the 
vacuum tube and is denned as 



_ dip 

Em . 

dv GK 



(4.72) 



Also, the parameter r v is known as the vacuum tube plate resistance and 
is defined as 



dv PK 
di PK 



(4.73) 



« c ^=constant 



The symbols and definitions of these parameters are standard in the 
vacuum tube industry. 

A second equivalent circuit can be found by applying Thevenin's 
Theorem to the circuit of Fig. 4.28. In this circuit, the current generator 
between the cathode K and plate P is converted to a voltage generator. 
The internal impedance of this voltage generator will be r P . (The value of 



Basic Amplifiers 



149 



internal impedance between terminals K and P when all current sources 
are open circuited and all voltages shorted.) In addition, the magnitude 
of the voltage source is equal to the open circuit voltage at the terminals 
K to P. For the generator of Fig. 4.28, this voltage is 

v = g m r v v glc (4.74) 



Let a new parameter /j. be defined as 

H = 



(4.75) 



where fi is known as the voltage amplification factor. With these changes, 
the equivalent circuit of a grounded grid amplifier can be represented as 



^~. 4. " p 



M"g* 



?G 



Fig. 4.29. A voltage source equivalent circuit for the tube of Fig. 4.27. 



shown in Fig. 4.29. This circuit is known as a voltage source equivalent 
circuit for the triode. From this circuit it can be seen that if the plate 
current is held constant, the plate-to-cathode voltage will change /jl times 
as much as the grid-to-cathode voltage. Accordingly, the parameter fi can 
be defined as 



dv rK 
/u = — — ! ^ 

dv UK 



(4.76) 



i /.=constant 



Note that ft is negative since v GK must increase in the negative direction 
as v PK increases in the positive direction in order to maintain i P 
constant. 

As with the transistors, the value of the vacuum triode parameters can 
be found from the characteristic curves of the triode. For example, the 
value of r„ is shown in Fig. 4.26 as the reciprocal of the slope of the line 
BC. Here the control grid is maintained constant at —4 v and the relation 
between v t > and i t , is established. For example, a change of 40 v in v P 
causes a change of 4.5 ma or 

40 



4.5 x 10- 



■ohms 



Hence, r P is equal to 8900 £2. In addition, the line AC represents a con- 
stant value of plate voltage (120 v). The relationship between v a and i F is 



150 



Electronic Engineering 



established to obtain an average value for g m . Av a is 2 v and Ai P is 4.5 ma. 
Hence, 

4.5 x 10" 3 



6m 



= 2250 micromhos. 



Also, the line AB represents a constant value of plate current (8 ma). 
Therefore the amplification factor may be obtained from the relationship 
between v P and v G along this line AB. In this example, &v G is — 2 v and 
Av P is 40 v. Therefore 

40 on 

u — = —20 

-2 
PROB. 4.15. Find the values of ft, r v , a.ndg m for the tube of Fig. 4.26 at the 
point where v P = 200 v, and v G = —6 v. Answer: ft = —19.6, r v = 8.5 KQ, 
g m = 2,300ft mhos. 

PROB. 4.16. Is r p the same for all points on the characteristic curves? What 
shape would the characteristic curves have if r v were constant ? 
PROB. 4.17. Is ft the same for all points on the characteristic curves? What 
condition must the characteristic curves meet if ft is constant ? 

4.9 THE GROUNDED GRID AMPLIFIER 

Just as the transistor can be used to provide power amplification, so 
is the triode tube capable of power amplification. As noted before, the 
circuit of the triode which conforms to the circuit of the common-bas e 
transistor is known as the grounded grid amplifier. The circuit of a typical 
grounded grid amplifier, together with the appropriate equivalent circuit, 
is shown in Fig. 4.30. 

From the equivalent circuit of Fig. 4.306, it is evident that i v = — i k when 
reverse (normal) bias is applied between the cathode and the grid. Thus, 
the forward current gain is unity with reverse bias, but less than unity if 
forward bias is used (positive grid). 







Vo 



(a) (b) 

Fig. 4.30. The circuit diagram and equivalent circuit of a grounded grid triode amplifier. 



Basic Amplifiers 

Since v f is v kg , which is — v gk , our loop equation is 

»<-/"»* = -'j.( r » + z i) 
(-/i + l)v t 



— J„ = 



»» = -«'j> Z £ = 



The voltage gain, 



Q _ V 0_ (-/" + ^ 



The input impedance, using Eq. 4.78, 

7 _»i vj_ _ r p + Z L 

The output impedance, from Fig. 4.31, 



Z. = * 



But -»,* = »*„ = 'pZ 9 



»«, = ijfv + Z g ) + f*v gk 

v b = i P (r p + Z g ) - fti p Z g 
Z„ = r p + (-fi + \)Z g 



151 

(4.77) 
(4.78) 

(4.79) 
(4.80) 

(4.81) 

(4.82) 
(4.83) 

(4.84) 
(4.85) 



It should be observed that the common grid connection has high voltage 
gain, low input impedance, and high output impedance. The impedances 



K 



Wgk 



?G 



P 

— o o 



«0 



Fig. 4.31. Circuit for determining output impedance of the grounded grid amplifier. 



152 



Electronic Engineering 



are affected by the load and source impedances in a manner similar to the 
common-base transistor. In fact, the grounded grid amplifier acts as an 
impedance transformer having an impedance transformation ratio of 
— n + 1 . The power gain for a resistive load is, 



G = 



ip'Rz. (-/* + 1)^ 



i„ Ri- 



r v + Rl 



For resistive loads, 



G. 



(4.86) 



(4.87) 



Thus, it becomes evident that che characteristics of a common-base 
transistor amplifier are similar to a common-grid vacuum tube amplifier. 

PROB. 4.18. Calculate the voltage gain, input impedance, and output imped- 
ance of an amplifier with R L = 50 KQ, R g = 10 KQ, r v = 8 Kfi, and,u = -20. 
PROB. 4.19. A transistor is connected as shown in Fig. 4.32. The voltages and 
currents that were measured are shown on the diagram. Draw an equivalent 
circuit using h parameters for this transistor. Give values of each component 
in the equivalent circuit. 



1.905 ma 



-1.865 ma 




lMamp 




Fig. 4.32. The configuration for Prob. 4.19. 



PROB. 4.20. A transistor amplifier is connected as shown in Fig. 4.33a. The 
transistor characteristics are given in Fig. 4.336. If v g = 2 sin cot, find: 

(a) The quiescent collector voltage and current. 

(b) The current amplification of the amplifier. 

(c) The voltage amplification of the amplifier. 

(d) The power amplification of the amplifier. 



Basic Amplifiers 



153 




-30 



12 3 4 5 6 

I E in milliamperes 
(b) 
Fig. 4.33. The information required for Prob. 4.20. (a) Circuit configuration; and 
(6) characteristics of the transistor in part a. 



5 



The Common Emitter 
Amplifier 



The reader may have surmised while reading Chapter 4 that a rearrange- 
ment of the input terminals of the common-base amplifier would reduce 
the input current requirement for a given output current and thus provide 
higher power gain. Such a configuration would employ the emitter as the 
common or grounded terminal, the base as the other input terminal, and 




Fig. 5.1. A common emitter amplifier. 
154 



The Common Emitter Amplifier 155 

the collector as the other output terminal. This type of circuit is known as 
the common emitter configuration. Figure 5.1 is a simple circuit diagram 
of this arrangement for a.p-n-p transistor. With the conventional current 
directions shown in this figure, i B , i c , and v are negative. Thus a polar- 
ity reversal occurs between the input and output signal voltages. 

5.1 CURRENT GAIN OF THE COMMON EMITTER AMPLIFIER 

In the common emitter configuration, the input current is the base 
current, i B . But 

Since the collector current i c is normally almost equal to the emitter 
current i E , the base current i B may be small in comparison with the collector 
current. Thus, current gain is achieved. Using Eq. 4.1 (reprinted below) 

— i c = «.i E — I co (4.1) 

- ig = kzJcg (52) 

oc 

Substituting this value of — i E into Eq. 5.1, 

i B = ^^ - ic (5-3) 

a 

o.i B = (1 — a)t' c — I co (5.4) 

Solving for i c explicitly, 

" in + ^ (5-5) 



" 1 - a 1 - a 

The term[ a/(l — a)|is called the sho rt circuit current jgain ox_current 
am £M£Sli ( ^If c f or f° r the commonemitter connection. This amplification 
factor is sometimes known as /S (beta). Other symbols which are commonly 
used in the literature are a. ct and h FK . Equation 5.5 may be rewritten: 

ic = #b + + Wco (5-6) 

PROB. 5.1. Prove that the coefficient of I co in Eq. 5.6 is correct. 
PROB. 5.2. A transistor has a = 0.99 and I co = 1 yua at a specified operating 
point. What value does p have? What will be the value of collector current 
when the base current is reduced to zero, assuming /? to be constant? Answer: 
fi = 99, i c =0.1 ma. 

One way to visualize the control action of the base is to assume the 
forward bias between the base and emitter to be increased. This increased 
bias causes the height of the potential hill, or barrier field, to be reduced. 



156 Electronic Engineering 

The injection current across the emitter-base junction is therefore increased. 
Most of the carriers which comprise this current, drift across the thin base 
region into the collector region. Thus, the increase in collector current may 
be much greater than the increase in base current. The collector current 
change is almost proportional to the base current change. Therefore, the 
base current is often assumed to be the control parameter. The reverse 
diode current I co is amplified because it is forced to flow across the base 

(0 + D/co 




Fig. 5.2. An illustration of the process by which I co is multiplied in the common 
emitter configuration when I B = 0. Actual current directions are shown. 

to emitter junction. As a result, a small forward bias is produced across 
that junction and the collector current is increased as though I co were 
added to the base current. The process by which I co is multiplied is 
shown in Fig. 5.2. 



5.2 CHARACTERISTIC CURVES 

A very useful set of curves is the family of collector characteristic, or 
i c vsv CE curves. A typical set oip-n-p collector characteristics is provided 
in Fig. 5.3. Comparison with the common base collector characteristics 
of Fig. 4.4 reveals that the following differences exist: 

1. The base current i B is a parameter in the common emitter set. 

2. The curves of Fig. 5.3 are not evenly spaced, which indicates that i c is not a 
linear function of i B . This should be expected since a is a function of i> as 
discussed in Section 4.4. 

3. The collector current is approximately equal to zero when v cv equals zero, 
regardless of the value of i B . The reason for this behavior is that'the collector- 
base junction has the same amount of forward bias voltage as the emitter-base 
junction when v CE equals zero. This condition may be verified by observing 
from Fig. 5.1 that v BC equals v CK in polarity as well as magnitude when v VF 
equals zero. The knee of each curve occurs when v CK is increased sufficiently 
to reduce the collector-base junction bias to zero. Larger values of v CF result 
in reverse bias. 

4. The spacing between curves increases as v cic increases because a and hence /? 
increase with v CE . 



The Common Emitter Amplifier 



157 



Because there are four independent transistor parameters, an additional 
set of curves is required. A set of input characteristics similar to the 
common base set of Fig. 4.5 is given in Fig. 5.4a. The curve obtained with 
v CE equals zero is widely spaced from the others because the base current 
is increased as a result of the carriers injected across the collector-base 
junction. The transistor does not normally operate in this region of very 
small collector voltage, which is known as the saturation region. Notice in 
Fig. 5.4a that v BE does not equal zero, generally, when i B equals zero. This 
small forward bias when i B equals zero results from I co flowing across the 




-10 -15 -20 -25 -30 

v CE in volts 

Fig. 5.3. Collector characteristics for the common emitter configuration. 



emitter-base junction, as explained previously. The emitter-base junction 
voltage could be reduced to zero by the application of a small reversed 
base current as indicated in Fig. 5.4a. This reversed base current would 
be nearly equal to I co as shown in Fig. 5Ab and the collector current is 
reduced to nearly I co . The base and collector currents differ somewhat 
from I C o because some of the carriers which comprise I co still (even 
with no bias) diffuse to the base-emitter junction where they are swept 
across this junction. 

The curves provided for the common emitter connection contain the 
same information as those for the common base configuration, because 
-i E == i c + i B . The common emitter curves are more useful, however, 
because i K may be readily and accurately obtained by the addition of the 
other two currents, but it is difficult to accurately determine i B from the 
common base curves because i B is the difference of two currents which are 
nearly equal. Therefore, the common emitter curves are used for all 
transistor configurations. 



158 



Electronic Engineering 



-0.25 



-0.20 



00 

3 



-0.15 



-0.10 



2 -0.05 




+20 



-20 -40 -60 -80 

Base current in microamperes 

(a) 



-100 



-120 




(b) 

Fig. 5.4. An illustration of the reduction of the base-emitter junction voltage to zero 
and the reduction of the collector current to I co by the application of a reverse base 
bias current equal to I co . Actual current directions are shown, (a) Common emitter 
input characteristics. (6) The circuit diagram. 




y bb v cc 

Fig. 5.5. Circuit for obtaining characteristic curve data. 



The Common Emitter Amplifier 159 

Figure 5.5 is a circuit which may be used to obtain data for the common 
emitter curves. The batteries and potentiometers may be replaced by 
variable voltage power supplies. The variable resistor R should be very 
large compared with the base-to-emitter resistance of the transistor so the 
base current will remain essentially constant while the collector voltage is 
varied. This large value of R requires a rather high base supply voltage. In 
addition, an electronic-type voltmeter must be used to measure v BE . The 
energizing current of a conventional meter may be larger than the base 
current and thus cause the reading of the meter which measures i B to be 
completely erroneous. Also, a highly sensitive voltmeter, preferably an 
electronic type, should be used to measure v CE . 

5.3 TRANSISTOR RATINGS 

The manufacturer specifies the maximum permissible collector voltage, 
collector current, and collector power dissipation of a transistor at a 
specified ambient temperature. This dissipation rating must be reduced as 
the ambient temperature increases. Under certain conditions, the dissipa- 
tion rating may be exceeded for very short periods of time, but until these 
conditions are thoroughly understood, it is only safe to assume that the 
transistor will be permanently damaged if any rating is exceeded at any 
time. Transistor damage may be prevented if the area of safe operation 
for the transistor is clearly indicated. For example, if the absolute maxi- 
mum ratings of the transistor whose collector characteristics are shown in 
Fig. 5.3 are; 

i* c max = —6 ma 

i; CE max = — 30 v 

collector dissipation = 35 mw at 25°C 

The dashed line on Fig. 5.3 marks the boundary of the area of safe 
operation. Before taking data for a characteristic curve plot, this area 
should be marked and kept vividly in mind so that no data points will be 
taken outside of the area of safe operation. 

The maximum collector voltage rating for the common emitter connec- 
tion may be considerably less than for the common base configuration. 
The reason for this reduced collector voltage rating may be seen from a 
study of Eq. 5.5 and Eq. 4.20. It should be observed that ($ approaches 
infinity as a approaches unity. Therefore, the collector current in the 
common emitter configuration will approach infinity, even with zero base 
current, as a approaches unity. 

PROB. 5.3. A transistor has a = 0.99 at low values of collector voltage. The 
avalanche voltage V A = 40 v. The exponent N = 3. Calculate the collector 



160 



Electronic Engineering 



voltage at which the collector current approaches infinity in the common 
emitter configuration. Hint: Using Eq. 4.20, assume that a = r\y at low values 
of collector voltage. Then, a = 1 when d = \\r\y. Answer: v c = 8.6 v. 

The avalanche breakdown voltage, listed as V A in Eq. 4.20, is commonly 
listed as V CBO in manufacturer's ratings : 

Vcbo ~ collector-to-emitter breakdown voltage with emitter circuit open 

The apparent breakdown voltage which occurs in the common emitter 
configuration when a = 1 and i B = is known as V CE0 . 

V CEO = collector-to-emitter breakdown voltage with base circuit open 

when V CBO and V CEO are known for a given transistor, V CBO is the highest 
and Vceo ' s the lowest collector breakdown voltage of this transistor. 




Fig. 5.6. Collector voltage breakdown characteristics and ratings for a typical transistor. 



Intermediate values of breakdown voltage occur when finite resistance 
is placed between the base and the emitter in the common emitter con- 
figuration. This breakdown voltage is known as V CKR at a specified 
base circuit resistance. 

V CRR = collector-to-emitter breakdown voltage with base 
resistance specified 

The highest V CER occurs when the base is shorted to the emitter. 
This value of V CER is known as V CKS . 

V CES = collector-to-emitter breakdown voltage with base shorted 
to emitter 



The Common Emitter Amplifier 



161 



The common emitter breakdown voltage is increased slightly above 
V C es an d is essentially equal to V CBO if reverse bias is applied between 
the base and emitter. This reverse-bias breakdown voltage is known as 
Vcex- 
V CEX = collector-to-emitter breakdown voltage with base reverse-biased 

The various breakdown voltages for a typical transistor are shown in 
Fig. 5.6. 

5.4 THE GROUNDED CATHODE AMPLIFIER 

The triode tube may be used in a configuration which corresponds to the 
common emitter configuration of the transistor. This circuit, known as a 
grounded cathode amplifier, is shown in Fig. 5.7. T he resistor R provide s 




Fig. 5.7. The grounded cathode triode amplifier. 

a conducting path be tween the grid and t he bias supply. This grounded 
cathode circuit is by tar the most widely used tube circuit. Consequently, 
the characteristic curves given by the manufacturers are for the grounded 
cathode circuit. 

If the control grid potential is maintained negative with respect to the 
cathode, the grid current is essentially equal to zero. Consequently, the 
vacuum tube is a voltage actuated device, whereas the transistor may be 
considered a current actuated device. A family of collector characteristics 
for a typical triode is shown in Fig. 5.8b. 



5.5 GRAPHICAL ANALYSIS 

A common emitter (or common cathode) amplifier may be designed with 
the aid of the characteristic curves in the same manner as that indicated 



162 



Electronic Engineering 




Rli for maximum 
power output 



Rl2 f° r ni B ner 
power gain and 
better linearity 



Collector volts 
(a) 



\ Rla for maximum 
\^ power output 




linearity but lower 

power gain and 

lower power output 



Plate volts 
(W 



Fig. 5.8. Illustration of graphical analysis by the use of load lines, (a) Transistor 
amplifier; (b) triode amplifier. 



for the common-base amplifier of Chapter 4 (Section 4.3). The following 
procedure might be used. 

1. Choose a value of collector supply voltage somewhat less than the maximum 
rating of the device, providing a resistive load is to be used. 

2. Draw a load line from the supply voltage (V cc or V 1>P ) through the safe 
operating range below the maximum dissipation curve so that the load line 
intersects the current axis below its maximum rating. See Fig. 5.8 for an 
example. The actual choice of load resistance will depend on the require- 
ments of the amplifier. Maximum power output is obtained when the 
maximum area falls below the load line, since power is the product of v and /'. 
Larger values of load resistance result in better linearity. Maximum power 
gain is obtained when the load resistance is equal to the internal resistance of 
the amplifier. These statements will become more meaningful as the work 
progresses. 

3. A quiescent operating point is selected in accordance with the requirements of 
the amplifier. If the input signals are symmetrical, an operating point near 
the center of the load line would provide maximum dynamic range. Such a 
point is selected for the examples of Fig. 5.9. 

The quiescent operating point of the transistor may be located on the 
input characteristic curves in the same manner as describle for the common 
base configuration. From these input characteristic curves the required 
forward bias voltage V SE may be obtained. However, the required bias 
current is usually obtained, by using a supply voltage which is large compared 
with V BE . The bias current is then controlled by a series-dropping resistor. 

The performance of the transistor amplifier can be determined by select- 
ing a convenient collector voltage swing about the quiescent operating 
point. The collector current swing and base current variation required to 
produce this output may then be determined from the load line on the 
collector characteristics, and the input characteristics may be used to 
determine the required input voltage variation. Thus the current gain, 



The Common Emitter Amplifier 



163 




•o 



* oj 



IX 

o 

u 
O 



J3 




O 

T3 
O 

■G 



so 
« 

o 

c 
o 



0\ 

vi 

oi 



164 Electronic Engineering 

voltage gain, and power gain may be determined as well as the magnitude 
of the permissible signal levels. The degree of nonlinearity may also be 
predicted for various signal levels and operating conditions. Similarly, 
the performance of a triode tube amplifier can be determined by selecting 
a convenient plate voltage swing about the quiescent operating point. The 
plate current swing and the grid voltage variation required to produce this 
output may be determined by traversing the load line on the plate charac- 
teristics. This load line method of determining the performance of an 
amplifier is illustrated in Fig. 5.9 for the common emitter configurations of 
both the n-p-n transistor and the tube amplifier. The voltage gain for the 
tube or the current gain for the transistor may be obtained by dividing 
the peak-to-peak output quantity by the peak-to-peak input quantity, as 
discussed in Chapter 4 in connection with the common base transistor 
amplifier. A qualitative measure of the amplifier distortion may be 
obtained by comparing the output waveform to the input waveform. A 
sinusoidal input waveform was assumed in Fig. 5.9 because the departure 
from sinusoidal waveform in the output is easily recognizable and measur- 
able. A quantitative discussion of distortion is delayed until Chapter 10. 

PROB. 5.4. Determine the approximate voltage gain of the triode amplifier 
whose characteristics, load resistance, and bias potentials are provided in Fig. 
5.96. What are the maximum permissible magnitudes of input and output 
voltage which may be obtained with reasonably good linearity ? 

The current gain of the transistor amplifier may be obtained from the 
collector characteristics and load line of Fig. 5.9a. However, an additional 
set of curves is needed for the graphical determination of the voltage gain 
of the transistor. As discussed in Chapter 4, a set of input characteristics 
may be used to obtain the needed relationship between the input current 
and the input voltage. A typical set of input characteristics for the common 
emitter configuration was shown in Fig. 5.4. In addition, an unusual 
set of input characteristics is given in Fig. 5.10. Also a set of collector 
characteristics for this hypothetical transistor is provided. From Fig. 5. 10 
it may be seen that the relationships between the base current and collector 
voltage taken along the load line are used to plot the dynamic input charac- 
teristic. In this figure the static input characteristic curves are quite widely 
and uniformly spaced. In contrast, the typical transistor may have the 
static input characteristics essentially merge into a single curve for all 
collector voltages above a few tenths of a volt. Also, the input charac- 
teristics may vary appreciably between individual transistors of a given 
type. Therefore, the manufacturer usually does not provide a complete 
set of input characteristics. Frequently the input resistance of the transistor 
is given as a function of the base current, and from this information the 
small signal voltage gain may be obtained. 



The Common Emitter Amplifier 



165 




-30-60-90 -120 -150 
Base current in microamperes 




-10 -20 

V CE in volts 



Fig. 5.10. Illustration of a method for determining the dynamic input characteristics 
of a common emitter transistor. 

PROB. 5.5. A symmetrical input signal is applied to the transistor of Fig. 5.10. 
Determine the current gain and voltage gain of the amplifier with the load 
resistance indicated when the quiescent operating point is at v CE = — 10 v and 
the base current variation is from —30 /na. to —90 fta. Answer: G 4 = 50, G v = 65. 

A second graphical method of solution quite widely used is by a graphical 
plot which relates the input voltage (or current as the case may be) to the 
output current (or voltage). This graphical plot is known as the dynamic 
transfer curve. Example 5.1 illustrates the procedure for finding the 
dynamic transfer curve and also demonstrates how the dynamic transfer 
curve is used to find the output current (or voltage). 

Example 5.1. A triode is connected as shown in Fig. 5.1 1 . Find i P if v z has the 
form shown. The characteristic curves of the 6J5 are shown in Fig. 5.12. 
Assume R L is 20,000 il. 



-3.0 

-4.5 

t/> 
"5 
» -6.0 

C 

:£" 

-7.5 
-9.0 






Time. 
'3 U 



«5 te 




Fig. 5.11. Circuit for Example 5.1. 




100 



400 



500 



200 300 

Plate voltage in volts 
Fig. 5.12. The plate characteristic of a 6J5 tube. (Courtesy of Radio Corporation of 
America.) 

Grid volts 
-18 -16 -14 -12 -10 -8 -6 -4 -2 



12 


















10 


Static curve — 






































§ 8 

CL 

E 
















X. 




















Bgt 


■-\ — 




c^_rr- lp 




2 6 














S\ 






— -f~ 4 i 1 


















1 






E 
2 4 

O- 












^— 


1 

" — V 


— 1-- 





rit&i 












^1 

| 




i 






1 i ! ! i 


1 
1 
1 


2 










L - 










h ti (3 u te 
Time >- 


<6 










1 
1 




i 






1 


i i ' 




1 
h r — 


J-'-Ia I 




u '2 | 

1 - ' 




"P 








13 T 

1 

<4 1 




V 







































Fig. 5.13. The dynamic transfer characteristic. 
166 



The Common Emitter Amplifier 167 

As the first step, the load line is drawn on the plate characteristics of the tube 
(Fig. 5.12). Values of input voltage (grid voltage) are known and values of 
output current (plate current) are required. Thus a plot of i P vs v G is the re- 
quired transfer curve. Since the tube will operate along the load line, the required 
values of i P and v can be taken from the load line. 

Hence, point x is chosen on the load line of Fig. 5.12. The value of grid 
voltage and the value of plate current corresponding to point x are noted. A 
point with the same value of grid voltage and plate current as point x is plotted 
on a new graph with v a as one axis and i P as the second axis. This new graph 
(with point x indicated) is shown in Fig. 5.13. 

This procedure of transferring points from Fig. 5.12 to Fig. 5.13 is continued 
until enough points are plotted in Fig. 5.13 to draw the dynamic transfer charac- 
teristic curve. Note that as R L approaches zero, the dynamic transfer curve 
approaches the static transfer curve. 

After the dynamic transfer characteristic has been drawn, the input signal vj 
is drawn as shown in Fig. 5.13. Values of voltage are then projected from the 
input signal (point A, Fig. 5.13) to the transfer curve (point B) and from the 
transfer curve to form the output signal (point C). 

The same process can be used for finding the dynamic transfer charac- 
teristics of a transistor. If the transfer of base current i B to collector 
current i c (or collector voltage v c ) is wanted, the load line values of either 
collector current or collector voltage may be obtained as a function of 
base current from the collector characteristics. For example, the dynamic 
transfer characteristics of i B vs i c are given in Fig. 5.146 for a load resist- 
ance of 1250 O. However, if the transfer of base voltage v B to collector 
current i c is required, both the input and the output characteristics of 
the transistor are required. For example, in Fig. 5.14c a base potential of 
—0.49 v is shown as point x. (The dynamic input characteristics are 
required for this step.) Point x must then be transferred to the corre- 
sponding base current and collector voltage of the collector characteristics. 
(See point x in Fig. 5.14a.) From these curves the required collector 
current can be read and the corresponding point plotted on the v B vs i c 
characteristics. Figure 5.14c? shows the total v B vs i c curve for the given 
load. 

The dynamic transfer characteristic can be used to check the linearity 
of an amplifier. If the input signal is applied to a straight portion of 
the dynamic transfer characteristic, the output signal will have the same 
shape as the input signal. If, however, a curved portion of the dynamic 
transfer characteristic is used, the output signal will have a different shape 
than the input signal. This change of shape is known as nonlinear dis- 
tortion. 

Figure 5.14 shows that the relation between base current and collector 
current is more nearly linear than the relation between base voltage and 



168 



Electronic Engineering 



« 



o S 

O Q- 

1 § 



i r 



i i i i 

s3J3duiB!||!iu ui jusjjno J0}oa||03 



vV 


\ u ^ 1 

\ \ 1 °| 
\ \ I tf) 1 

\ l\ l\ ' 1 






"A 


L\ 1 


















«o 


\\ 

C3 \ 

c* \ 
in I 

CM 1 










ii \ 





I 



i a 





<0 








n 










> 










CM 
1 


o 





I I I I 

saj3dLue|||;iu u; }uajjno jopanoQ 



The Common Emitter Amplifier 



169 



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170 » 



Electronic Engineering 



collector current. Hence, the transistor is best used as a current-activated 
device. 

PROB. 5.6. Draw the dynamic transfer characteristic for the tube used in 
Example 5.1, but with the load resistance R L increased to 50,000 0. Which 
value of R L gives the greater linearity ? 

PROB. 5.7. The transistor of Fig. 5.14 is connected to a 1000 €l load resistor. 
The voltage source V cc is —5 v. Plot v c if i B is —150 + 100 sin mt //a. 
PROB. 5.8. Draw the dynamic characteristic of i c vs i B for the transistor of 
Fig. 5.14 if R L is 2000 CI. Compare this curve with Fig. 5.146. 

5.6 EQUIVALENT CIRCUITS FOR THE COMMON EMITTER 
TRANSISTOR 

The h parameters for the common emitter transistor configuration may 
be obtained from the collector and input characteristics of Fig. 5.3 and 




-20 



-40 -60 -80 

Base current in microamperes 

Fig. 5.15. Curve for obtaining h f , 



-100 



-120 



Fig. 5.4 in the same manner as described in Chapter 4 for the common-base 
configuration. For example, the slope of a collector characteristic (v C e vs 
i c curve with i B constant) would be 

u -Ale 



dv, 



CE 



(5.7) 



i R =coi)8tant 



This parameter is the output admittance with the input open. If the output 
current i c is plotted as a function of the input current i B for a constant 
value of collector voltage, the slope of this curve is the forward current 
transfer ratio 

di f 



h, e = 



c 
din 



(5.8) 



i' f , £ =constiint 

This parameter is the dynamic or small signal value of /?, the current 
amplification factor of the common emitter configuration. The static or 



The Common Emitter Amplifier 



171 



d-c value of (i is frequently indicated by the symbol h FE . Figure 5.15 
is a plot of i c as a function of i B for the transistor of Fig. 5.3 at v CE = 10 v. 
The slope of this curve at a given point is h u at that point. Note that in 
contrast to h th (which is negative) h u is positive. 

The slope of the static input characteristic is the input impedance with 
the output shorted. 

dv B E 



K. 



dh 



(5.9) 



v c£ =coaBiAnt 



The input characteristics (Fig. 5.4) could be used to plot a family of 
v BE vs v CE curves with i B as the constant parameter. This plot is made in 



a 



E 



-0.30 
-0.25 
-0.20 
-0.15 
-0.10 
-0.05 



"0 -5 -10 -15 -20 -25 

Collector-to-emitter voltage in volts 

Fig. 5.16. Curves for obtaining h re . 

Fig. 5.16. The slope of any one of these curves at a particular point is the 
reverse voltage transfer ratio at that point. 





_100(ia 










-60^ 










_40t*q_ 










-20|ia 










j l = j -1C 


(^ 






•* ■" 











_dVjBE 



dv, 



CE 



(5.10) 



inconstant 



The term h re is also known as the reverse voltage amplification factor fi cb . 

As previously mentioned, h re and h te may be determined by taking the 
ratio of incremental values of the pertinent variables about the operating 
point. Hence, it is not necessary to plot the curves of Figs. 5.15 and 5.16. 

The differential equations which describe the operation of the transistor 
are, 



dv 



&VBE = 



\s -{He 

din 



BE 



M B + 



dv 



'BE 



dv, 



CE 



A», 



CE 



i B =k 



a;« + 



v CE =k 



di c 
dv, 



CE 



A», 



CE 



(5.11a) 



(5.11b) 



is=k 



172 



Electronic Engineering 



From the defining equations for the preceding h parameters (Eqs. 5.7 to 
5.10) and the symbolism Av BE = v oe etc., the common emitter /j-parameter 
circuit equations for the time-varying components of voltage and current 
can be written. 

V» = Keh + he v ce 
h = h nh + h oeV C e 

The common emitter A-parameter equivalent circuit is presented in Fig. 
5.17. In solving this A-parameter circuit, it is important to remember that 
when a resistive load is connected to the output terminals, the voltage v ce 
is negative. 



(5.12) 





Fig. 5.17. Common emitter A-parameter equivalent circuit. 



The solution of a common-emitter A-parameter circuit is easily carried 
out by assuming that the input current i b is known. This method of 
solution is illustrated with the aid of Fig. 5.18, where the transistor load 
resistance is R L . The output voltage 



»„. = 



where g L = ljR L . 



-h fe i b 
he + Sl 



Vbe = hKe + hreVce 

Substitution of Eq. 5.13 into Eq. 5.14 yields 



The input resistance is 



= hKe - 



t>be 



Kehfeh 
hoe + 8L 



in h 



Kehfe 
,e + gL 



(5.13) 



(5.14) 



(5.15) 



(5.16) 



Note that except for the second subscripts, this equation (5.16) is the 
same as Eq. 4.64. In fact, all of the common emitter A-parameter 
equations have the same form as the corresponding common base 
/j-parameter equations. 



The Common Emitter Amplifier 

The voltage gain is (using Eqs. 5.13 and 5.15) 



G _ Vce _ -h, e 

v be h ie (h oe + g L ) - h re h fe 



173 



(5.17) 



Note that G v is a negative quantity which indicates a polarity reversal 
between output and input voltages. (The denominator will always be 
positive since h ie h oe is greater than h re h fe .) 



lb 


>- 






■< 


-lc 


4 


' ^ 1 






C 


i 


Vbe 


1+ 


( 


7\hfeib * 


>hoe * 


>R L 


c 


hreV<x\ ) 
) 1 


— o 






o 



Vce 



Fig. 5.18. An ^-parameter circuit for a common emitter transistor with a load 
resistance R L . 



PROB. 5.9. Using the A-parameter circuit, solve for the current gain of the 

common emitter transistor in terms of the //-parameters and the load resistance. 

PROB. 5.10. Solve for the output resistance of the common emitter amplifier in 

terms of the A-parameters and the driving source resistance R s . 

PROB. 5.11. Determine the common emitter h parameters for the transistor 

whose characteristics are shown in Fig. 5.14. Use the quiescent operating 

point indicated as x. Answer: h ie - = 800 fl, h re = 1 x 10^ 2 , h fe = 15, h oe = 

2 x 10-* mhos. 

PROB. 5.12. Using the /t-parameters found in Prob. 5.11, calculate the current 

gain, voltage gain, and power gain of the transistor with a load resistance of 

1.25 Kft. 

The two-generator A-parameter circuit is convenient for solving any 
common-emitter circuit problem. The one-generator equivalent T circuit 
will be developed, however, in order to provide a better understanding of 
the common-emitter configuration. 

Figure 5.19a is the common emitter arrangement of the equivalent T 
circuit of Fig. 4.20. Inspection of the circuit will show that the input 
impedance with the output open (z u ) is r b + r e . Figure 5.196 illustrates 
the method of determining the input impedance when the output is 
shorted, (h ie ). 

v be = h r b + (h + Qr e (5.18) 



174 

but i e = h u i h when the output is shorted. Then 

"be = V» + Oft + Kh)r e 
and (with the output shorted) 

2*! = h ie = r 6 + (1 + /!,>. 



Electronic Engineering 

(5.19) 

(5.20) 



Inspection of Eq. 5.20 and Eq. 4.59 reveals that the input impedance with 
the output shorted is h u + 1 times as large as the input impedance of the 
common-base circuit under these conditions. Then 



h ie = (hfe + Wib- 



l b > _ + Jk^ _ + 



(5.21) 



Tmf>e 



Vbe 



r m^e 






ie]] 



r mle 



Vce 



(a) (b) (c) 

Fig. 5.19. An equivalent T circuit for the common emitter configuration. 

PROB. 5.13. Prove Eq. 5.21 is valid and explain why the input impedance of 
the common emitter configuration is increased by this factor (h u + 1). 

Figure 5. 19c illustrates the method of determining the output impedance 
when the input is open, l/Ao«- 



Vce ^c'c 'e e ' mse 

but since i b = 0, —i e = i„. Then 

Vce = *'oOc + r e — r m ) 
and the output impedance with the input open is 

From the relationship r m — a.'r c , 

Z„ = (1 - a')r e + r e 

•Z22 == " = r d "I" r e 
hoe 

Where r d = (1 — a')^ by definition. Then r d = rj(h u + 1). 



(5.22) 

(5.23) 

(5.24) 

(5.25) 
(5.26) 



The Common Emitter Amplifier 

Since r a » r e and r c » r b 

u l 1 



*/.+ ! 



: (h u + l)h ob 



175 



(5.27) 



^22 r d r c 

Thus it is seen that when the input is open, the output admittance of the 
common emitter configuration is (h u + 1) times that of the common 
base. 

An equivalent T circuit which would give a more obvious indication of 
the output impedance and also give the output current and voltage in 
terms of the input current would be a definite improvement over the 



o-WW- 



ffib 

■e- 



■WW- 

rd 



r b rd ^~f 



r m ib 



(a) (b) 

Fig. 5.20. Improved equivalent T circuits for the common emitter configuration. 

equivalent T circuits of Fig. 5.19. Two circuits incorporating these advan- 
tages are given in Fig. 5.20. It is quite evident that the current generator 
of Fig. 5.20a would give the proper short-circuit collector current. From 
this circuit the open circuit generator voltage is 



tfgen = fi'h r d 
a' 



f gen — 



rj h 



(5.28) 
(5.29) 



1 - a 

t>gen = a'r c i 6 (5.30) 

tfgen = rj b (5.31) 

This r Rcn is the same as the open-circuit voltage of the generator of 
Fig. 5.20Z>. It is also the same open-circuit voltage as that provided 
by the generator of Fig. 5.19 because i c = and i b = /, when the output 
is open. 

PROB. 5.14. Prove that the short-circuit currents of the generators of Fig. 5.20a, 
Fig. 5.20b, and Fig. 5.19 are identical. 
PROB. 5.15. Prove thut h fe = -(h fe + l)h, b . 



176 Electronic Engineering 

In the preceding development, all of the h parameters have been found 
in terms of the r parameters except h re = fi cb . This parameter is the ratio 
of the voltage which appears at the open circuit input terminals to the 
voltage v ee which produced this input terminal voltage. From Fig. 5.21, 

v ee = Ur 4 + r e ) (5.32) 

V be = l >« 

h re = ^ = l — = r ' (5.33) 

v ce i c (r„ + r e ) r d + r e 

The h parameters for both common base and common emitter con- 
figurations have been determined in terms of the r parameters. Conversely, 
the r parameters can be determined in terms of the h parameters as follows. 

o — M/VW 1 — n/WWV (J ? 

Vce 



o 

Fig. 5.21. Circuit for determining h„ = fi cb . 

The value of r d + r e in Eq. 5.26 is equal to l//i oe . When this term is 
inserted into Eq. 5.33, we find that r e is 

(5.34) 
(5.35) 

(5.36) 

(5.37) 
*/. + 1 
Also, all of the common emitter h parameters except h re have been deter- 
mined in terms of the common base h parameters. A summary of these 







r e = — 










hoe 






From Eqs 


. 5.20 and 5.34 
r 


= h ie -(l+h fe )^ 
hoe 






From Eq. 


5.26, r d +r e = 


\lh M , and using Eq. 5.34 
r 1 - h re 

h oe 






We note in Eq. 5.5 that /3 


or h fe is equal to a/(l — 


a). 


Thus 






h " 







The Common Emitter Amplifier 

relationships follow. 

hie = (hfe + i)h ib 

h oe = (h u + l)h ob 
From Prob. 5.15, h fe is 

K - -{h u + Vj-hf,, 

Finally, from Eqs. 4.61, 5.35, and 5.27, h re is 

"re = "tt"ob "rb 



177 

(5.21) 
(5.27) 

(5.38) 
(5.39) 



PROB. 5.16. Find the common emitter h parameters for the transistor of Table 

4.1. 

PROB. 5.17. Verify Eq. 5.39. 

The h parameters are most often given by the manufacturer. Since the 
h parameters as well as the r parameters vary with the terminal voltages 
and currents of the transistor, the parameters are normally specified or 
obtained at some specific operating point. In addition, a plot of the 



10 

5 
» 

! 2 

» 

! i 

> 

[0.5 

I 

0.2 




1 2 5 10 20 50 100 
V C B~\n volts 

(a) 



0.1 





hu, 








v°y 












\.p, 




hrb 








hfe- 












hob 


V 








hib 















0.1 0.2 0.5 1 2 5 
7^ in milliamperes 

(b) 



10 



Fig. 5.22. A plot of the variation of the h parameters as a function of (a) collector to 
base volts ; (b) emitter current. 



parameters as a function of the terminal voltages and currents is sometimes 
given as shown in Fig. 5.22. This presentation makes it possible to obtain 
the parameters quickly at any other operating point. The parameter 
values are normalized to values listed by the manufacturer at the given 
operating point, that is, the point i E = 1 ma and v CB = 5 v in Fig. 5.22. 

PROB. 5.18. A 2N527 transistor has h u = 81, h ie = 2.4 K ohms, h oe = 30/t 
mhos, and h re = 0.79 x 10 -4 . Find the r parameters and also the common 
base h parameters for this transistor. 



178 



Electronic Engineering 



PROB. 5.19. Assume the h parameters given in Prob. 5.18 are valid for v CB = 5 v 
and i E = 1 ma. If the h parameters vary as given in Fig. 5.22, find the h param- 
eters for v CB = 10 v, and i E — 2 ma. Answer: h u = 128, h ib = 18.1 Q., h ob = 
0.43 fi mhos, h rb = 6.5 x ICT*. 

5.7 THE EQUIVALENT CIRCUIT OF A COMMON 
CATHODE AMPLIFIER 

A common cathode amplifier circuit is shown in Fig. 5.23a. The 
equivalent circuit for this configuration is given in Fig. 5.236. This 
equivalent circuit can be obtained by merely rearranging the equivalent 




(a) 




(b) 
Fig. 5.23. A common cathode amplifier, (a) The actual circuit, (b) the equivalent 
circuit. 

circuit for the grounded grid triode (Fig. 4.28). However, this circuit can 
also be found by the differential equation for plate current. 






diji 



Av, 



(5.40) 



A AV ° + A 

where i P is the plate current, v u is the control grid voltage, and v P is 
the plate voltage. In terms of small-signal components, this equation can 
be rewritten as 



g m v 9 + — v P 



(5.41) 



The Common Emitter Amplifier 



179 



where /„, v g , and v p are the time-varying components of i P , v G , and v P , 
respectively and r„ and g m are as denned in Eqs. 4.73 and 4.72. Accord- 
ingly, an equivalent y-parameter circuit for the triode is shown in Fig. 
5.23b. 

If the control grid is maintained negative with respect to the cathode, 
essentially no current flows in the control grid circuit. Hence, the control 
grid is drawn in Fig. 5.236 as an open circuit. The foregoing equations and 
terms are exactly the same as those in Chapter 4 (Section 4). This similarity 
should have been expected since the derivation in Chapter 4 used the 
cathode as the reference electrode and the circuit just described also uses 





K 



Fig. 5.24. An equivalent circuit containing a voltage generator. 

the cathode as the reference electrode. This similarity can be carried one 
step further by replacing the current generator of Fig. 5.236 by a voltage 
generator. As in Chapter 4, Thevenin's theorem is used in this transition. 
The resulting circuit is shown in Fig. 5.24. The current generator has 
been replaced by a voltage generator with an open circuit potential p 
times as great as the grid potential. As in Chapter 4, [i is negative and 
is given by Eq. 4.75 

/" = -g m r, (4.75) 

Of course, the values of r p , /i, and g m can be found from the characteristic 
curves as explained in Chapter 4. 

The usual type of characteristic curves for a tube have the form shown 
in Chapter 4 (Fig. 4.26). However, other configurations may be used. For 
example, if the plate current is plotted as a function of grid voltage with 
the plate voltage held constant, the slope of the curve at any point is the 
transconductance at that point. This curve is known as a static transfer 
characteristic curve, as previously mentioned. A family of transfer 
characteristics is given in Fig. 5.25, where a curve is obtained for each 
different value of plate voltage. 

Since the characteristic curves are not straight lines, the parameters of 
the tube (ft, g m , and r v ) must not be constant. The typical variation of tube 
parameters with plate current is shown in Fig. 5.26. This figure indicates 
why the tube parameters must be found for the region in which a given 



180 



Electronic Engineering 

25 



-9 




-8 



-7 



-6 



-5 -4 -3 

Grid voltage in volts 



-2 



-1 



Fig. 5.25. Transfer characteristics for the 12AT7 triode. 



tube will be operated. The same tube operating in a different region will 
have different parameters. 

Equations can be derived 1 for the fi of a tube, and they indicate that fi 
is determined by the physical structure of the tube. The change of n is 
due to variations in physical spacing of the elements and the edge effects 
of the elements. However, a similar equation for the g m of the tube 



-22 

4-20 

-18 






_M 
























f&m 








30,000 
| 20,000 






































o 














,?• 10,000 




















T~ 




















3000 



2000 



E 
a 



4 



1000 



2 4 6 8 10 12 

Plate current in milliamperes 

Fig. 5.26. Curves showing the dependence of ft, g m , and r„ on the operating point of 
a 6J5 triode. 

1 For development of these equations, see Karl Spangenberg, Vacuum Tubes, McGraw- 
Hill, New York, 1958, pp. 125-128 and pp. 135-137. 



The Common Emitter Amplifier 



181 



indicates g m is dependent not only on the physical structure of the tube 
but also on the current density in the tube. Hence, the great variation in 
g m with plate current in Fig. 5.26 results. Since the plate resistance is 
equal to fj. divided by g m , r p must behave as illustrated in Fig. 5.26. 

PROB. 5.20. Find the values of /i, g m , and r „ for the tube whose characteristic 
curves are given in Figs. 5.25 and 5.27. Use the operating (£))point of v G = — 1 v 
and v P = 175 v. 




100 



200 300 400 

Plate voltage in volts 



500 



600 



Fig. 5.27. Characteristic curves of a 12AT7 triode. (Courtesy of Radio Corporation 
of America.) 



PROB. 5.21. Draw two equivalent circuits for the 12AT7 tube of Prob. 5.20. 

The equivalent circuits of Fig. 5.23 or Fig. 5.24 may be used to determine 
the magnitude of the input and output impedances of the common cathode 
tube amplifier. Assuming the grid does not draw current, the input 
impedance of the tube itself is 

Z in = oo (5.42) 

In a later section, it will be shown that an input capacitive term does 
exist. However, for low frequencies (less than hundreds of kilohertz) 
the input impedance of the tube may be considered as essentially infinite. 
There must be a d-c conducting path between the grid and the bias source, 
however, to maintain the desired bias on the grid. Therefore, the input 
impedance to the amplifier circuit is dependent on the resistance of this 
d-c conducting path and effective input capacitance of the tube and not 
on the load resistance. This behavior is in contrast with the transistor 
amplifier. 



1*2 Electronic Engineering 

Because Z 12 is zero, there is no reverse amplification factor for the triode 
tube. Therefore, in contrast with the transistor, the output impedance of a 
tube is not a function of the driving source impedance. It may be seen 
from the equivalent circuits that 

^out = r, (5.43) 

Consequently, the triode tube amplifier circuits are easier to analyze than 
the transistor circuits because of the rather complete isolation of input 
and output circuits. 

Perhaps the most interesting behavior of the triode tube is the current 
gain. From Fig. 5.23 and Fig. 5.24, it is obvious that the current gain of 
the tube itself is essentially infinite. This interesting fact may be verified by 
referring to Eq. 5.5. 

*/. = r 5- (5 - 5) 

1 — a 

Since a is essentially one for a tube with a negative grid, h fe = oo for the 
tube. Also, since the power gain is equal to the voltage gain multiplied by 
the current gain, the power gain is essentially infinite at low frequencies. 
Of course, this power gain does not consider the signal power which may 
be consumed in the grid circuit resistor. When the grid resistor is taken 
into account, the tube amplifier has the same order of power gain as the 
transistor amplifier. 

We can find the voltage gain of a triode amplifier by using the equivalent 
circuit. For example, the circuit of an actual amplifier and an equivalent 
circuit of this amplifier are shown in Fig. 5.28. The capacitor C„ is very 
large and therefore blocks the d-c bias voltage due to V nn but passesj he 
t ime-varying signal v t . Therefore, the signal component v g of the tota l 
grid voltage v G i s equal to v t . A loop equation can be written for the 
plate circuit. 

-W- = f,(r„ + R L ) (5.44) 



or 



<>„ + R L ) 



The output voltage is 



v = -i p R L 
for the direction of current and v shown. Thus, 



-£fc 



The Common Emitter .Amplifier 

The total voltage gain is 



f*Ri 



° V v t (r, + R L ) 



183 



(5.47) 



Since fi is negative the output voltage is inverted in relation to the 
input voltage. This inversion is characteristic of all grounded emitter 
circuits. Also, from Eq. 5.47, the voltage gain of the circuit increases as 
R L increases. However, the maximum voltage gain of the circuit in 
Fig. 5.28 will never be greater than fi. 




(b) 
Fig. 5.28. A common cathode amplifier, (a) Actual circuit; (6) equivalent circuit. 



PROB. 5.22. The plate characteristics of a 6J5 tube are shown in Fig. 5.29. If 
the 6J5 is connected as shown in Fig. 5.28, find the voltage gain. V PP is 300 v 
and V GG is -6 v. The value of R L is 50,000 CI. Answer: G„ ~ -13. 
PROB. 5.23. Repeat Prob. 5.22 if R^ is changed to 20,000 CI. Compare the 
voltage gain and amplifier linearity for these two values of load resistance. 



5.8 BIASING CIRCUITS FOR THE TRANSISTOR 

The circuit diagrams used in the preceding discussion employ two bias 
batteries. In the common emitter transistor configuration the batteries 
have the same polarity with respect to the common electrode. Therefore, 
a single battery or d-c power source may be used as illustrated in Fig. 5.30 
to provide the biases. The potential difference between the base and the 
emitter is normally very small in comparison with the collector supply 



184 



Electronic Engineering 



16 



12 







o/ 


























II 1 

s7 


?/ 
























1 


















*/ / 






'/ 




















/ 






/ 


71 
















~/ 


/ 






h 


t 


o# 


■^n. 












1 


/ 






' 1 






'/ 












/ 


/ , 












/ 










/ 


I / 








f 


■a 






































*C< 


£ 


y 


) 100 




200 


300 


400 50 










PI 


ste volta 


?e 


in vol 


s 











Fig. 5.29. Plate characteristic curves of a 6J5. (Courtesy of Radio Corporation of 
America.) 

voltage V cc . Therefore, nearly all of V cc is dropped across R f and the 
base bias current is 



/n~ 



R, 



(5.48) 



The value of R f may be calculated after proper values of V cc and I s have 
been determined. This type of bias is known as fixed bias because the base 
bias current does not vary with the transistor parameters. 




Fig. 5.30. Fixed bias. 



The Common Emitter Amplifier I* 5 

Fixed bias is satisfactory providing the transistor parameters are 
accurately known and the temperature remains fairly constant. If the 
parameters change because of temperature change or substitution of 
transistors, the operating point will move along the load line as shown in 
Fig. 5.31. The initial design placed the quiescent operating point in the 



5 

w A 
<u 

CX 

£ 3 
.2 

1 2 
1 




5 

IA A 

a. 

E 3 
.2 

I 2 
Z 














^ — 120 /ua 


s_ - 


■100 


80 






^ 






— 60 




f~ ^ 


s. 






' 


s 






-40 




V 


:*. 


20 








X 







120 jua 


/^""" 


-inn 




80 

* 60 


~-\ 






\,_^ on 




\ 


> 


\«b = 




i B = 



(b) 



(0 



Fig. 5.31. Shift in quiescent operating point when fixed bias is used. 



center of the load line, as indicated in Fig. 5.31a. An increased temperature 
causes the saturation current to increase and the characteristic curves to 
shift upward as shown in Fig. 5.316. This moves the quiescent operating 
point upward on the load line. If a transistor with a lower current 
amplification factor were substituted, the characteristic curves would be 
compressed, as shown in Fig. 5.31c. The quiescent operating point would 
then move downward along the load line. Conversely, an increased h u 
would cause the operating point to move upward. 

Several methods are used to stabilize the operating point. In the simplest 
method known as self bias, the base bias current I B is derived from the 



186 



Electronic Engineering 



collector potential v CE rather than from the collector supply potential 
V cc . This method of bias is illustrated in Fig. 5.32. Any change in 
parameters which would cause an increase in collector current would 
decrease the collector potential because of the IR drop across the load 
resistor. This decrease of collector potential would decrease the bias 
current, which in turn would tend to restore the collector current and 
voltage toward the initial operating point. The collector voltage at the 



i — ^^w 




x c 



Hl- 




JL 



(a) 



(b) 



Fig. 5.32. Self-bias. 



quiescent operating point, V CE , is usually large in comparison with the 
base-to-emitter voltage V BE . Therefore, 



R f 



'CIS 



(5.49) 



PROB. 5.24. Calculate values of bias resistance R f for both fixed and self bias 
for the transistor amplifier whose characteristics and operating conditions are 
given in Fig. 5.14. Answer: Fixed bias 33.3 KCi, self bias 13 Kfl. 

The degree of stabilization realized with self bias depends on the magni- 
tude of the ohmic resistance of the load. In some cases a transformer is 
used to couple the signal components in the collector circuit to the load. 
In this case the ohmic resistance in the collector circuit is small, the 
collector voltage is almost equal to the supply voltage, and self bias would 
not provide much better stability than fixed bias. Another problem arising 
from self bias is that the signal (time-varying) components of collector 
voltage cause signal currents to flow through the bias resistor R t . These 
currents are of opposite polarity to the input currents and tend to cancel 
them. This effect decreases the gain of the amplifier and is called defenera - 
tion or negative feedback. This degeneration can be eliminated bv dividing 
the bias resistance R f into two parts as shown in Fig. 5.326 and using the 



The Common Emitter Amplifier 



187 



part which is attached to the collector in conjunction with a capacitor C 
to filter out the signal components. This method is not an ideal arrange- 
ment, however, because R n loads the input and R n loads the output of the 
transistor. 

Another factor which needs to be considered in the stability problem is 
the variation of base-to-emitter voltage with temperature, for a given 
value of emitter current. The diode equation developed in Chapter 3, 
when applied to the emitter base junction becomes 



V r („QVBE/kT 1\ 



(5.50) 



where I EO is the saturation current of the emitter-base junction. But I EO 
varies widely with temperature as discussed in Section 3.4 and illustrated 




veb 



Fig. 5.33. Variation of v EB or i E with temperature. 



in Fig. 3.6. Typical i E vs v EB curves are given in Fig. 5.33 for three 
different temperatures. Assume that T is the ambient temperature and 
that the bias point is O. If the emitter current is held constant, the 
emitter-to-base voltage will decrease to A of Fig. 5.33 as the temperature 
is raised to T x or increase to B as the temperature is lowered to T 3 . On 
the other hand, if the base bias voltage were to be held constant, the emitter 
current would vary from C to D as the temperature varied from T x to T 3 . 
This would be the condition if battery bias were used with no series 
resistance. This situation could be approximated when transformer 
coupling is used. In case the bias source is a large voltage in comparison 
with v BE and a resistance is used to control the bias current, as was done 
in the fixed and self bias circuits, the emitter-base junction voltage may vary 
with temperature and the emitter current is not affected appreciably by 
the variation of v EB with temperature. 



1" Electronic Engineering 

Differentiation of the diode equation will show that the voltage across 
a junction decreases approximately 2.2 mv per °C temperature increase 
if the current across the junction remains constant. This variation of 
junction voltage with temperature holds for silicon as well as for ger- 
manium transistors. 

The effects of emitter and base circuit resistances on collector current 
thermal stability is shown in Fig. 5.34. In Fig. 5.34a, the emitter resistance 
is infinite and, therefore, all of I co flows through the base circuit resistor 
R B and none across the emitter-base junction. Therefore, the thermal 




(fi + l)ho 



*)0 + IMco 




=: rb 




Fig. 5.34. The effects of base and emitter circuit resistances on the collector current 
thermal stability. 

component of collector current is I co . In Fig. 5.34*, the base circuit is 
open and all of I co is forced to flow across the emitter-base junction and 
is amplified as shown in Fig. 5.2 and by Eq. 5.6. Therefore, the thermal 
component of collector current is (fi + \)I co . In Fig. 5.34c, both R B and 
R E are finite, so a fraction k of I co flows through R B and the remainder 
(1 — k)I co flows across the base-emitter junction and is amplified by fi as 
shown. The fraction k depends on the ratio of R K to R B . 

A commonly used collector current stability factor S T (actually an 
instability factor) is defined as the ratio of the change in collector current 
(A/ c ) to the change in T co (M co ) or, considering incremental changes, 



S/ = 



JIc 
dl 



CO 



all bias values held constant 



(5.51) 



The circuits of Fig. 5.34 provide stability factors (a) I, (b) fi + I, and 
(c) (1 — fc))S + 1, respectively, as may be observed. Note that (a) and 
(b) are the limiting values of the general case (c). 

A practical transistor bias circuit must provide forward bias to establish 
the desired quiescent operating point in addition to providing suitable 



The Common Emitter Amplifier 189 

thermal stability to maintain the Q point within a desired or specified 
range. A circuit that will accomplish these purposes is shown in Fig. 
5.35. This circuit is known as a stabilized bias circuit. In order to deter- 
mine the current stability factor a Thevenin's equivalent circuit is drawn 




(a) (b) 

Fig. 5.35. (a) A stabilized bias circuit, (b) Thevenin's equivalent circuit for (a). 

for the stabilized bias circuit as shown to the left of the points A and B 
in Fig. 5.35a. It may be seen by inspection of Fig. 5.35a that 



and 



R B = 



F n = 



■R1R2 
RiVcc 



(5.52) 



(5.53) 



Rt + Ri 
Kirchhoff's voltage equation may be written for the bias circuit of Fig. 



5.356. 



I e Re + v kb ~ JbRb ~V b = 



(5.54) 



But the current stability factor involves I c and I co , not I E and I B . 
Therefore, the following relationships must be used: 



-I K = I c + h 



Jn = 



Ir + 



'CO 



1 — a 1 — a 

Then, from Eq. 5.5, 

B \ 1 - a/ a a a 



(5.55) 
(5.5) 

(5.56) 



l9 " Electronic Engineering 

Substitution of Eq. 5.56 and Eq. 5.55 into Eq. 5.54 yields 

a a ' \ a a / 

(5-57) 
Kearranging and collecting terms, we find that 

'°[0 + ^K + l -r*.] - Mv +*)-*+ "- 

w , • ,. (5.58) 

Multiplication of both sides of Eq. 5.58 by a, gives 

We + (1 - a)Je B ] = / co (/? £ + R B ) + (F EB - V B )« (5.59) 
The current stability S 7 = dI c [dI co can be determined from Eq. 5.59 if 
V EB and a are assumed to be constant. However, variations of V EB and 
a with temperature, which may be very important in some circuits, need 
to be included in the total stability problem. This more general solution 
may be obtained if Eq. 5.59 is rearranged as follows. 

**I0 ~ «)'c - Ico\ = ReVco ~ /<?) + ( v eb ~ V B )x (5.60) 
Using the approximations 1 — a ~ 1/0 and a ~ 1, 

Rb\~ - Ico) ^ ReQco ~ Ic) + V EB - V B (5.61) 

The objective of the stabilized bias circuit is to confine the Q-point 
excursion to a predetermined range on the d-c load line for a specified 
range of ambient temperature, as shown in Fig. 5.36. 

In order to illustrate the increase of with temperature which occurs 
in some transistors, particularly the silicon type, collector characteristics 
are given for the maximum and minimum transistor temperatures. The 
high temperature curves are not required if a curve of h FE (or 0) is given 
as a function of temperature. I C1 and / C2 are the lower and upper limits 
of the @-point collector current as set by the design engineer. These 
permissible current extremes are set so that clipping does not occur in 
the output in any prescribed condition of operation. 

Assuming that the values of I C1 and / C2 are known, that A eoi and I C02 
are the corresponding values of thermal current at 7\ and T 2 , and that 
0! and /? 2 are the corresponding values of ft at these temperatures, Eq. 
5.61 can be written twice— once with values substituted for temperature 
7\ and again for values substituted for temperature T z . The first of these 
equations can then be subtracted from the second with the following 
result. 



rJIs* - in 

' fit fir 



Ico) = R E 



A/fo) = R K ^'co - A/ c ) + kV K „ (5.62) 



The Common Emitter Amplifier 



191 



I Cl anuLiv EB = y EBi 



and AK RR = V Kn , — V EB1 . 
(5.63) 



where A/ co = I C02 — I COl , M c = I C2 

Then, 

R^Mco - A/q) + ^Vrb 
K B = — ^^_^_^^— .^^— ^— ^^~ 

IC2 Ici A J 

— ~ — - M C o 
Pi Px 

Observe that R B may be determined as a function of R B , or vice versa. 
Therefore, an appropriate value must be chosen for either R E or R B and 





Fig. 5.36. Curves used to illustrate the determination of a stabilized bias circuit, 
(a) Collector characteristics at temperature 7\; (b) Collector characteristics at tempera- 
ture T 2 . 

then the value of the other resistance may be determined from Eq. 5.63. 
A suitable value for R E would be a value such that the voltage drop across 
it, I K R K , is large compared with AV BE but small compared with V cc . 
Values of I E R E between one and three volts are typical. An example 
will be given to illustrate a method of determining the bias circuit com- 
ponents for a stabilized bias circuit. 

Example 5.2. A germanium transistor is to be used in an R-C coupled audio 
frequency amplifier circuit which must furnish 1.0 ma peak signal current to a 
3.3 KO load. The amplifier must operate properly over an ambient tempera- 
ture range of 25"C to 50°C. The collector supply voltage V cc = -20 volts. 

We will first draw a 3.3 K load line on the collector characteristics as shown 
in Fig. 5.37a by the dashed line. But the total d-c load resistance includes the 
emitter circuit resistance R K , which must, therefore, be chosen before an accurate 
d-c load line can be drawn. If the low temperature Q point is tentatively chosen 
at BcK = -12 v and I c = —2.4 ma, I K would also be approximately 2.4 ma. 
Now let us select 1 K R E = 2.0 v, and the desired value of R K = 820 O. An 
accurate d-c load line, the solid one, can now be drawn (Fig. 5.37a) and the low 



192 



Electronic Engineering 



temperature Q points Q x set at V CE = -12 v, I C1 = -1.94 ma. This collector 
current is high enough to permit the required 1 ma peak signal current without 
clipping. The maximum temperature Q point can now be determined. First, 
note from Fig. 5.37a that the collector current cannot exceed 4.5 ma for this 
load line. Therefore in order to avoid clipping of the one ma signal at the maxi- 
mum temperature this Q point collector current I Ci must not exceed 3.5 ma. 
Thus, A/ c = l C2 -I C1 = -1.56 ma. 

In Chapter 3, Eq. 3.31, the thermal current of a germanium diode at 300°K 
was found to double for each 10°C increment of temperature. This rule will be 
used to determine A/ co . However, the junction temperature limits must first be 




Ambient temperature, *C 
(b) 



-25 



Fig. 5.37. (a) Collector characteristics used to illustrate the g-point excursion, (b) 
Derating curve used to determine thermal resistance. 

determined. The junction temperature is hotter than the ambient temperature 
because a thermal resistance T exists between the junction and the ambient 
surroundings. The difference in junction and ambient temperature is equal to 
the power dissipated at the junction times the thermal resistance. 

The derating factor (equal to l/d T ) is usually given by the transistor manufac- 
turer for air-cooled transistors, but sometimes a derating curve, as shown in Fig. 
536b, is given instead. Either the derating factor or the thermal resistance fly may 
be determined easily from the derating curve. The derating factor is the slope of 
the derating curve &PJ&T in watts (or milliwatts) per °C and the thermal resist- 
ance is the reciprocal of this slope, or &T/AP d in °C per watt (or milliwatt). The 
slope of the derating curve is the maximum dissipation rating P a of the transistor, 
usually given at 25°C, divided by the temperature difference AT between this 
rating temperature and the maximum permissible junction temperature. In the 
example under consideration, the derating factor is 200 mw/60°C = 3.33 mw/°C 
and the thermal resistance T = 1/3.33 = 0.3°C per milliwatt. 



The Common Emitter Amplifier I' 3 

The junction temperature limits can now be determined. The minimum 
junction temperature is 

T A + P d T = T A + (.v CQ1 i CQ1 )8 T = 25°C + (23.3X0.3) =* 32°C 

at the lowest G-point limit. The maximum junction temperature is 

50°C + (3.5 x 5.58)(0.3) = 56°C 

Now, I co must be determined at these temperature limits. 

The manufacturer usually gives maximum and typical values of I co at 25°C. 
The value of I co for a particular transistor can, of course, be readily measured. 
The maximum I co at 25°C for the given transistor is listed as -12 /w, which is 
a safe design value for any transistor of this type. The following relationship will 
determine I co at any junction temperature when a reference I c0 is given. 

Tj-T 

I co (T) = I co (T )2 T * (5.64) 

where T is the reference temperature, T i is the junction temperature, and r d is the 
temperature difference over which I co doubles. Thus, in this example 

I co min = (-12 iid)i Ao = -19.5 na 

and 

/ co max =(-12/ia)2 3 ^» = -103 fia 

Then, 

A/ co = -83.5 //aand5 7 = A/ c /A/ C0 = -1.56/0.0835 = 18.7. 

As previously stated, the emitter-base voltage V EB decreases approximately 
2.2 millivolts per °C temperature increase. Since AT = 56°C — 32°C = 24°C in 
this example, AV BB = 24(-2.2 x 10" 3 ) = -0.053. However, this decrease is 
partially offset by the increased emitter current at the higher temperature. In 
this example, the partial compensation due to emitter current change will be 
neglected. 

The jS of a germanium transistor varies mildly with temperature and therefore 
will be considered constant at a value of 90 over this small temperature range. 

The value of R B can now be calculated by using Eq. 5.63. 



* R = 



820[-83.5 x 10" 6 -(-1.56 x 10~ 3 )] - 0.053 



B 1 Sfi x 10~ 3 

~ %0 -(-83.5X10-*) 

_82 0(1.47 x 10- 3 ) - 0.053 1.205 - 0.053 _ 

B -1.73 x 10- 5 + 83.5 x 10- 6 66 x 10" 6 

The resistors R x and R 2 (Fig. 5.35) must be determined in a manner that 
will provide the desired Q-point values. This may be accomplished by the 



M Electronic Engineering 

simultaneous solution of Eqs. 5.52 and 5.53, repeated here for convenience. 



R i R i 



V * = R^k (5 " 53) 



Solving Eq. 5.53 for R 2 yields the relationship 

*i(Ycc - V B ) 



R, = 



(5.66) 



Substituting this value of i? 2 into Eq. 5.52 and solving for R lt we have 

V CC R B 

and finally 



/ 
Ri = y _ v (5.67) 

V CC V B 



*2 = -^ (5.68) 

The voltage V B must now be determined. Equation 5.54 can be rearranged to 
yield 

v b = I E Re + V EB - I B R B (5.69) 

If the input characteristics for the transistor are available, V EB can be accurately 
determined at the Q point. If these characteristics are not available, an ade- 
quately accurate estimation may be made. For example, V EB may be estimated 
at 0.2 v for germanium and 0.5 v for silicon in small signal amplifiers. Using 
this estimate in our example, 

V B = 2 + 0.2 + (45 x 10- 6 )(17.4 x 10 3 ) = 2.98 v 

the Q-point base current I BQ (which is equal to I B in Eq. 5.69) can be deter- 
mined as -45 (ia either from the collector characteristics in Fig. 5.36a or from 
Icol h FE< providing that h FE at the Q point is known. The values of R 1 and R 2 
can now be determined for our example : 

Rj. = 17.4AT(20)/(20 - 2.98) = 20.2 Kfl 

R 2 = 17.4A:(20)/2.98 = 117 KQ 
This completes the design of the bias circuit. 
PROB. 5.25. Verify Eqs. 5.66, 5.67, and 5.68. 

In this example, the emitter circuit resistor R E was chosen so that I E R E 
was large in comparison with AV EB . Consequently, the AV EB term in 
Eq. 5.63 can be dropped with little error, as illustrated by Eq. 5.65. Also, 
as assumed here, /S is almost independent of temperature (over moderate 
ranges) in a germanium transistor. With these constraints, Eq. 5.63 can be 



The Common Emitter Amplifier 



195 



simplified as follows. 



Rr = 



Ale 



(5.70) 



-A/ 



CO 



Multiplying both numerator and denominator by /?, dividing both numera- 
tor and denominator by A/ co , and using A/ c /A/ C0 = S r , Eq. 5.70 
becomes 

fi R E (1 - Sj) WASj - 1) 
i?»~ = KP-' 1 ) 

Sj-fS p-s, 

and whenever 1 « S x « ft, a further simplification can be made. Then 

R B ~ SjR E (5.72) 

PROB. 5.26. A given transistor which has h u = 100 is used in a stabilized 
bias circuit which has R E = 1 KO. The required stability factor S t is 10. 
Determine the base circuit resistance R B by both the method of Eq. 5.71 
and the approximate method (Eq. 5.72) and compare the values. Repeat the 
calculations for Sj = 5 and Sj = 20. 

The resistor R E will seriously reduce the gain of the amplifier unless a 
bypass capacitor is connected in parallel with R E as shown in Fig. 5.38a. 
The voltage gain of the amplifier without the bypass capacitor is approxi- 
mately equal to the ratio of the collector circuit resistance R c to the emitter 
circuit resistance R E . Since the voltage i E R E is large compared with 
V EB , the input voltage is almost equal to i E R E and the output voltage 
is equal to i c R c - Then, since i c ~ i E , the voltage gain is G v ~ R c IRe- 




i— O — VvV- 1 — Wv— *'> 



Re > dfzCg 




(a) 



(b) 



Fig. 5.38. (a) Stabilized bias including bypass capacitor, (b) Equivalent circuit as seen 
by C E . 



1"» Electronic Engineering 

The stabilized amplifier essentially has full power gain over the range 
of frequencies for which the bypass capacitor is effective. The lowest 
frequency at which the bypass capacitance is considered effective is the 
frequency at which the magnitude of the capacitive reactance is equal to 
the resistance which is being bypassed. This frequency is called the low 
cutoff frequency / x (or <o x in radians per second). The capacitor C E 
appears to be bypassing R E , but actually R E is in parallel with the transis- 
tor resistance as seen from the emitter terminal. In fact, this transistor 
resistance is usually so much smaller than R E that R E can be neglected. 

The transistor resistance as seen from the emitter terminal can be 
determined with the aid of Fig. 5Mb. In practical circuits, R c is very 
small in comparison with r c and the desired resistance is the same as the 
input resistance of a common-base amplifier with output shorted (h ib ) 
except that the parallel combination of R B a nd the driving source resistance 
R s appear in series with r h . Let this parallel combination be defined as 
R' s , assuming that the reactance of the coupling capacitance C c is negli- 
gible compared with R s at the lowest frequency of interest. With these 
assumptions, the required bypass capacitance is determined to be 

c ^ ri'Vk, (5 - 73) 

Full power gain is achieved for the amplifier at frequencies 10 w x or 
higher. Thus, if frequencies down to w 1 are to be amplified with essentially 
no loss of gain, a capacitor ten times as large as that given by Eq. 5.73 
will be required. 

PROB. 5.27. Derive Eq. 5.73. 

The chief disadvantage of the stabilized bias circuit, aside from its 
complexity, is the shunting effect that the bias resistance R B may have 
on the input signal in the event that R B is not large in comparison with 
the input resistance of the transistor. There are several advantages of 
the circuit, however, that make this circuit attractive. In addition to 
stabilizing the Q point of a given transistor from the standpoint of 
temperature variation, the stabilized bias circuit permits interchanging 
transistors that have appreciably different parameters without altering 
the bias circuit (see Fig. 5.31 for the effect of varying h, e ). This feature 
makes assembly line amplifier production possible without the necessity 
of individual tailoring of bias components or the requirement of excessively 
close tolerance limits for the transistors involved. Also, the signal shunting 
effect of R B , which was mentioned as the chief disadvantage, may actually 
be an advantage in a mass produced amplifier because it tends to stabilize 
the amplifier gain when transistors of varying h u are used. This feature 



The Common Emitter Amplifier 197 

results from the fact that transistors with high h u also have high input 
resistance {h ie = (h u + l)h ib ), and hence the shunting effect of R B appre- 
ciably reduces the gain of only the higher gain transistors. Silicon 
transistors often have such small values of I co that thermal stabilization 
is not required; however, stabilized bias is usually used in order to gain 
the other benefits mentioned. 

Measured values of I co (listed by manufacturers) include the surface 
leakage component. Therefore, the rate of I co increase with temperature 
is usually less than the values derived in Chapter 3. Although germanium 
transistors follow quite closely the rule of doubling I co every 10°C, silicon 
transistors, because of their relatively small ratio of thermal current to 
surface leakage current, usually have a much slower rate of increase than 
the theoretical doubling every 6 degrees. Manufacturers data should be 
consulted, especially in the case of silicon transistors. 

PROB. 5.28. A 2N2712 n-p-n silicon transistor has the following character- 
istics: h, e = 200, maximum collector dissipation = 200 mw at 25 °C, maximum 
junction temperature = 100°C, I co = 0.5 fia at 25°C. I co doubles for each 
15°C temperature increase. 

This transistor is used as a common emitter amplifier with V cc = 20 v and 
R L = 10 KQ. The Q point (at 25°C) is at V CE = 10 v and I c = 1 ma. V SE = 
0.5 v at this Q point. Draw the diagram of a stabilized bias circuit and determine 
the value of all resistors which will permit a 1.0 v change in V CE when the 
ambient temperature is raised from 25 to 85 °C. Allow a 2.0 v drop across the 
emitter resistor. Use any valid approximations. 

PROB. 5.29. Calculate the emitter bypass capacitance required for the amplifier 
of Prob. 5.28 if the desired lower half-power frequency is 15 Hz (o> x = 100 rps) 
and the driving source resistance (excluding the bias circuit) is 5 Kii. Verify the 
validity of the assumptions made in Prob. 5.27. Use h ie = 5 KSi. 

5.9 CATHODE OR SELF BIAS FOR TUBES 

The bias battery of a triode amplifier can be eliminated if cathode bias 
is used. The circuit which accomplishes this desirable simplification is 
shown in Fig. 5.39. T he average d-c plate current, I P , flowing through R K 
produces a voltage drop with the polarity shown. Since the cathode is 
positive with respect to ground and the control grid is at ground potentia l, 
the grid is negative with respect to the cathode. T his type of bias is also 
jmown as xelfhias. 

The ca pacitor C^ is placed in the cathode circuit to bypass the time- 
varying signal to ground. If the resistor R K is not bypassed, degeneration 
of the time-varying signal results. T his degeneration can be visualized by 
observing that an increase in control grid potential causes an increase in 
ca_thode _current. If C K is not present, this increase of cathode current 



198 



Electronic Engineering 



causes an increase of voltage across R IZ . As a result, the total potential 
b etween grid and cathode is not as great asThis potentialwouldbe if th e 
voltage across R K were constant. The _capacitor C K provides a low 
imEJdance jjath to ground for the time^varying signal . T herefore the 
magnitude of the time-varying voltage across R K is very smalL " 



e a -p,»w. 




Fig. 5.39. A cathode-biased amplifier. 
The desired value of R K can be found by applying Ohm's law. 



G 



P _ « 
K K = — 



(5.74) 



where V a is the quiescent bias voltage desired and I P is the quiescent plate 
current. Both of these values can be obtained from the load line on the 
characteristic curves. 

The capacitor C K must bypass R K . However, as was indicated in the 
transistor configuration just considered, the input impedance of the tube 
(as viewed from the cathode) is also in parallel with R K . As given in 
Eq. 4.81, the input impedance to the cathode is 

Z« = r ~ £± £f (4.81) 

Thus, if the reactance of the capacitance, C K , at w, is to be negligible 
(say iV or less) in comparison with R K , the following relationship must 
hold. 

\Q{R K + Z t ) 



C K > 

If Z, » Rk> tn ' s equation reduces to the simple form : 

10 



C K > 



o^Rx 



(5.75) 



(5.76) 



The Common Emitter Amplifier 199 

The bias circuit for the triode is much simpler than some of the stabilizing 
circuits required for a transistor. However, since the current through a 
tube is not affected by the temperature of the tube, the problem of 
stabilization does not enter. As an added advantage, the cathode bias 
circuit is self stabilizing. If plate current should tend to increase through 
the tube, the voltage drop across R K would increase. This increased 
voltage drop causes the negative bias on the tube to increase, which 
tends to decrease the plate current. 

PROB. 5.30. (a) Find the value of R K which would be required in Example 5.1 
if self bias is used instead of fixed bias. (6) Assume the amplifier must pass fre- 
quencies as low as 60 Hz, what size C K should be used? (c) Draw the self 
bias circuit and give the proper values of components. Assume V G (at the Q 
point) is -6 v. Answer: (a) 1.05 KQ, (b) C K = 45 fifd. 

PROB. 5.31. Design a 6J5 amplifier stage which will give a voltage gain of 15. 
Use a V PP supply of +300 v and operate the tube at a quiescent control grid 
potential of -4 v. Use a cathode bias circuit. List the value of all resistors and 
capacitors if the lowest frequency to be amplified is 50 Hz. 
PROB. 5.32. Using the curves of Fig. 5.3 and Fig. 5.4, select an operating 
point for R L = 10 Kft and determine the current, voltage, and power gain of 
the amplifier. 

PROB. 5.33. A 6J5 tube is connected as shown in Fig. 5.11. (The characteristic 
curves are given in Fig. 5.12.) If the quiescent grid voltage is —6 v, and R L = 
30 Kfl determine the current gain, voltage gain, and power gain of the amplifier. 
(Use R g of 500 Kfl) 

PROB. 5.34. Draw the dynamic characteristics for v c vs v B for the transistor of 
Fig. 5.14. Assume R L is 1500 fi and V cc is -5 v. 
PROB. 5.35. Repeat Prob. 5.6 but use the dynamic transfer curve. 
PROB. 5.36. Using an equivalent T circuit, derive an expression for the input 
impedance of a common emitter transistor which has a load resistance R L in its 
output. 

PROB. 5.37. A circuit is connected as shown in Fig. 5.11. If R L is 30 Kii and 
v z is ( -4 + 0.002 sin tot), what is the voltage output? Which method would be 
the most accurate in this problem, load line, transfer characteristic, or equivalent 
circuit ? 

PROB. 5.38. A given silicon transistor has h fe = 100, maximum collector 
dissipation = 200 mw at 25°C, maximum junction temperature = 175°C, and 
I co = 2.0 pa at 75 °C. This transistor is used in a circuit with V cc = 30 v and 
a nominal Q point at V CE = 10 v and I c = 2 ma. The environment of the 
amplifier requires an ambient temperature variation between 50°C and 100°C. 
Design a stabilized bias network which will restrain the Q point range from 
V CE = 9 v to V CE = 11 v. Assume that I co follows the theoretical variation 
of I co with temperature. 

PROB. 5.39. A given type transistor has a range of h fe between 50 and 200. 
An amplifier manufacturer wishes to use these transistors in a circuit with 
V cc = 25 v and a nominal Q point at V CE = 10 v and I c = 3.0 ma. Design 



200 



Electronic Engineering 



a stabilized bias circuit that will confine the Q point between V CE = 9 v and 
v cb = 1 1 v, assuming that the junction temperature remains constant. 
PROB. 5.40. Under certain conditions, it may be rather difficult to measure 
the voltage transfer ratio h r and the current transfer ratio h f . This statement 
is particularly true at high frequencies where not only a magnitude but also a 
phase angle must be included. Fortunately, it is possible to determine all four 
h parameters by making simple impedance or admittance measurements with 
a bridge. (An r-f bridge is required at high frequencies.) One set of circuit 
configurations which can be used is shown in Fig. 5.40 (the circuits are a-c). If 
the circuit represents a transistor, d-c bias would have to be supplied with 
appropriate blocking capacitors to isolate the a-c and d-c components. 






Fig. 5.40. Configurations which can be used to determine the h parameters by use of 
a bridge. 

With a transistor connected as shown in Fig. 5.40, the following impedance 
and admittances are measured on an r-f bridge (Fig. 5.40): 



Z x = oXK)^ 25 ^ ohms 
Y 3 = (6.8 x 10- 2 ) mhos 



Y 2 = (5 x lO- 4 )^ 



mhos 



(8.08 x lO- 4 )^ mhos 



Find the h parameters for this transistor. {Hint: Note that in (a) V 2 = in 
(b) I x = 0, in (c) V 1 = V 2 , in (d) V 1 = 0.) 

PROB. 5.41. A radio set manufacturer has purchased a large quantity of npn 
silicon transistors for use in automobile radios. The variation of /S of these 
transistors is from 50 to 200 and the maximum value of I co is 0.5 //a at 25°C. 
I co doubles, approximately, for each 10°C temperature increase. 

The audio-frequency amplifier in the radio has V cc = 13 volts, R c = 4.7 KC1. 
The maximum expected ambient temperature range is from — 30°C to 75°C. 
Design a bias circuit which will confine the Q point within the limits v CK = 5 
volts and o CE = 8 volts over this range with random selection of the transistors. 
Assume v BE = 0.5 volt at 25°C. Select R E and determine the base circuit 
resistors. (Common-emitter configuration.) T 0.75°C per mw. 



6 



Multielectrode Tubes 
and Transistors 



The triode tube was invented in 1906, but it was the early 1930's before 
anyone attempted to place more than one grid between the plate and 
cathode. 1 The first multigrid tube was the tetrode, which contains two 
grids in addition to the cathode and plate. The physical arrangement of 
a tetrode is shown in Fig. 6. 1 . This tube is a triode with an additional grid, 
placed between the control grid and the plate. This second grid is called 
a screen grid because it acts as an electrostatic shield for the plate. 

6.1 TETRODE TUBES 

The need for an additional grid became evident when amplification of 
radio frequencies was attempted with triode tubes. A certain amount of 
capacitance (usually in the order of several pf) exists between the plate 
and grid of a triode. Consequently, when the triode amplifier is designed 

1 Encyclopedia Britannica, Vol. 8, page 34QH gives a short history of the development 
of the vacuum tube. 

201 



202 



Electronic Engineering 



to produce high gain at high frequencies (frequencies above a fraction of a 
megahertz) this grid to plate capacitance is large enough to couple a 
significant amount of power from the plate circuit back into the grid 
circuit. As will be discussed later, this feedback of signal may cause the 
amplifier to oscillate and thus become useless as an amplifier. Conse- 
quently, elaborate circuits were designed to counteract the effects of the 
grid-plate capacitance. With the introduction of the screen grid, the 
interelectrode capacitance at lower radio frequencies ceased to be a 



Filament 




Fig. 6.1. The physical arrangement of a tetrode. 

problem. The capacitance from the plate to grid of a conventional tetrode 
is of the order of a fraction of a pf. 

The symbol for a tetrode is shown in Fig. 6.2. The general pattern of 
the symbol is evident. 

The screen grid could be connected to ground or cathode and would 
serve as a good electrostatic shield between the plate and control grid. 
However, with the control grid negative and the screen grid at zero 
potential, the plate would be unable to attract an appreciable number of 
electrons. The plate current would, therefore, be very low. By maintaining 
the screen grid at a constant positive potential, the electrostatic shielding 
effect between the plate and control grid is maintained, but electrons from 
the cathode are attracted toward the positive screen grid. When the 
electrons arrive at the screen location some of these electrons are inter- 
cepted by the screen wires and become screen current. However, most 
of these electrons pass between the screen wires and continue on to the 
plate of the tube. For reasons to be explained later, the potential of the 



Multielectrode Tubes and Transistors 



203 



Control grid 




Screen grid 



Cathode 



Filament 



Fig. 6.2. The graphical symbol of a tetrode. 

screen grid is usually lower than the potential of the plate. A typical 
circuit arrangement of a tetrode amplifier as well as typical values of 
supply voltages are shown in Fig. 6.3. 

The characteristic curves for a type 24A tetrode are shown in Fig. 6.4. 
Notice that a change of plate voltage (above 90 v) causes very little change 
in the plate current. In effect, the electric field in the cathode-control grid 
region is shielded from the plate by the screen grid. Therefore, the value 
of r v for a tetrode is very high. In fact, the plate resistance of the type 
24A tube shown in Fig. 6.4 is about 0.5 megohm. 

The unusual dips in the characteristic curves for plate voltages between 
8 and 85 v require further explanation. When the curve of screen current 
vs plate voltage is displayed as well as the curve of plate current vs plate 
voltage, the reason for this unusual behavior becomes clear. Figure 6.5 
illustrates the two curves as well as a curve showing the variation of total 
cathode current (plate current plus screen current) as a function of plate 
voltage. The total cathode current is seen to be almost independent of the 
plate voltage. However, as the plate voltage is increased above 8 v the 
plate current begins to decrease, because the kinetic energy required of a 




Fig. 6.3. Typical tetrode configuration. 



204 



Electronic Engineering 



8.0 

03 6.0 

a. 

| 5.0 

I 40 
.E 3.0 

£ 2.0 
1 1.0 

I ° 

°"-1.0 

-2.0 





1 


+ 75 v 
































lfc=2.5y 
























flu 
























f 












- vei » • 










































l 


t 














-1.5v 




















1 


s 














1 1 
-3.0 V 




















1 , 


*■ 














— 1 — 1 
-4.5 




















// 


'A 


































| \— 






S3 


^ 


^ 








tt< 


^ 
















-6.0v 








^ 


^ 


1 — 


# 


*** 





































































20 



40 



60 



80 100 120 140 160 180 200 
Plate voltage in volts 

Fig. 6.4. Characteristic curves of a type 24A tetrode. 

primary electron is about 8 to 10 ev (depending on the type of metal) in 
order to release a secondary electron from a metal sheet. Therefore, as 
the energy of the primary electrons reaches about 8 ev, some secondary 
electrons are emitted from the plate. 

These secondary electrons are released in the region between the plate 
and the screen grid. Since the screen grid is more positive than the plate, 
the electrons are attracted to the screen grid. The net current to the plate 
is equal to the primary current minus the secondary current. Therefore, 
the total current to the plate decreases. At the same time, the current 
to the screen grid increases due to the additional secondary electrons. As 
the energy of the primary electrons increases, the number of secondary 
electrons also increases. In fact, as the curves of Fig. 6.4 indicate, when 
the plate voltage is between 15 and 60 v, the number of secondary electrons 




8v 15v 60v 85v 

Plate volts 

Fig. 6.5. Plate current and screen current characteristics for a tetrode. 



Multielectrode Tubes and Transistors 205 

may be greater than the number of primary electrons. The plate then acts 
as an electron emitter, and the plate current is reversed. As the plate 
potential becomes more positive than the screen potential, some of the 
secondary electrons are attracted back to the plate. When the plate 
voltage is 85 v or greater, essentially all of the secondary electrons are 
returned to the plate.* If the secondary electrons were not present, the 
characteristic curve would have the shape indicated by the dotted curve 
of Fig. 6.5. 

The characteristic curve indicates that the tetrode has a negative plate 
resistance for low values of plate voltage. (An increase of plate voltage 
causes a decrease in plate current.) This negative resistance characteristic 
of the tetrode, like the tunnel diode, has been used to produce oscillations. 
Oscillators will be discussed in subsequent chapters. However, for most 
amplifier applications the dip in the characteristic curve is undesirable. 
This dip seriously reduces the useful operating range of the amplifier. 
Consequently, the regular tetrode has been largely replaced by the pentode 
or beam power tetrode which is discussed in Sections 6.2 and 6.4 of this 
chapter. 

PROB. 6.1. Sketch v as a function of time if the tube of Fig. 6.4 is connected 
as shown in Fig. 6.3. R L = 30 Kft and v t = —4 sin cot. Would you recommend 
operation of this tube under these conditions ? Why ? 

PROB. 6.2. Draw the dynamic transfer characteristic for the tube whose 
characteristic curves are given in Fig. 6.4 if (a) R L = 40 Kfl, (b) R L = 40 KCi, 
( c ) R L = 100 Kii. Assume the tube is connected as shown in Fig. 6.3. 

6.2 PENTODE TUBES 

The undesirable secondary emission effects of the tetrode led to the 
development of the pentode tube. The pentode, which has three grids 
between the cathode and the plate is essentially a tetrode with a grid 
inserted between the screen grid and the plate. This third grid is called 
the suppressor grid. The physical arrangement and symbol for the pentode 
tube are shown in Fig. 6.6. 

The suppressor grid is usually connected to the cathode. A basic 
pentode amplifier circuit is shown in Fig. 6.7. As in the tetrode, the field 
from the positive screen grid causes electrons to move from the cathode 
toward the screen grid. The screen grid intercepts some of these electrons 
but most of them continue on toward the plate. The suppressor grid 

2 About 90% of the secondary electrons are emitted with an energy of 10 ev or less. 
All of the electrons are emitted at different angles. As a result, the component of energy 
away from the plate is usually less than the net energy of the electron. Consequently, 
very few electrons have a net energy away from the plate greater than 10 ev. 



206 



Electronic Engineering 



Plate 




Fig. 6.6. The pentode construction and graphical symbol. 

causes a field of low potential to exist between the screen grid and the 
plate. However, the electrons which were not captured by the screen grid 
have enough energy to pass through this low-potential field to the plate. 
A few electrons which travel straight toward the wire of the suppressor 
grid will either be returned to the screen grid or captured by the suppressor, 
depending on the energy they possess. 

The electrons which arrive at the plate cause the emission of secondary 
electrons, as in the tetrode. However, the suppressor grid is negative in 
relation to the plate and returns the secondary electrons to the plate. 
The suppressor grid, therefore, does not suppress the emission of secondary 
electrons but does suppress the effect of the secondary emission. 

The characteristic curves of a typical pentode are shown in Fig. 6.8. 
In this figure, the dip due to the secondary emission is absent. Except for 
this difference, the curves are very similar to the tetrode curves. The 




Fig. 6.7. A basic pentode amplifier circuit. 



Multielectrode Tubes and Transistors 



207 



advantage of the pentode over the tetrode is obvious from the charac- 
teristic curve. The potential on the plate of the tetrode had to remain at 
least 10 v higher than the screen grid potential or the characteristic curves 
weri nonlinear. In contrast, the plate of the pentode can be reduced far 
belcjw the screen-grid potential and the operation is still on the linear 



14 



12 



•10 



/ 


<r 








- 


1.5 








I 


o 


—4 — 
















I 













- 


1 
















V 








Gri( 


i No. 1 \ 


/olts*fc = 


= -1.5 






If 
















-2 




r 
















2.5 




is- 


















-3 


' s — 




















' 


















-4 


^ 





















100 



400 



500 



200 300 

' "' Plate voltage in volts 

Fig. 6.8. Characteristic curves of a 6AU6 pentode (K„ 2 = 150 v). (Courtesy of Radio 
Corporation of America.) 

portion of the characteristic curve. If the plate has a very low voltage, a 
cloud of electrons will accumulate near the suppressor grid. Conse- 
quently, a virtual cathode forms in this region, and the characteristic curves 
for very low plate voltages have the same form as the vacuum diode. As 
the voltage increases, all the electrons in the screen to plate area are 
attracted to the plate and a saturation condition exists in which the plate 
potential has a very small effect on plate current. Therefore, the plate 
resistance of a pentode is very high. 



6.3 EQUIVALENT CIRCUITS FOR PENTODES 

Equivalent circuits can be drawn to represent the tetrode or the pentode. 
Since the pentode contains one more grid than the tetrode, the equivalent 
circuit for the pentode will be found and modifications which will fit this- 
development to the tetrode should be obvious. 



208 Electronic Engineering 

The pentode contains three grids and a plate. Consequently, if the 
cathode is taken as reference, each of these electrodes should have an 
effect on the plate current. Then, for small signals, the law of super- 
position can be used to determine the characteristic equation for the 
pentode current. 



A, P = gm A» Gi + g m 2 At> G2 + g n3 Av G3 + — Av 



(6.1) 



where g m indicates the effectiveness of v ai in controlling the plate current. 
The parameter g m can be measured by applying proper potentials to the 
various grids and plate. Then, note the change of i P as v G1 is changed 
slightly. All other grid and plate potentials must be maintained constant 
while this change of plate current and grid 1 voltage occurs. Mathe- 
matically, this relationship may be written with the help of partial 
derivatives as 

g m = |^- (6.2) 

dv G1 

Similarly, the rest of the parameters in Eq. 6.1 can be written as follows: 

Sm* = ~ (6.3) 



H 



g mS = 1^ (6.4) 



G3 



r v = -^ (6.5) 

In a typical pentode amplifier circuit, the screen voltage v Gi and the 
suppressor voltage v G3 are maintained constant. Consequently, Av Gi = 
Av G3 = and Eq. 6. 1 can be written 3 as 

A 'p = g m Ai> ei + — Av P (6.6) 

r p 

This equation is identical to the characteristic equation of the triode. 
Consequently, the equivalent circuits for the pentode, for the tetrode, and 
for the triode are the same. For convenience, the two equivalent circuits 
for the pentode or tetrode are shown in Fig. 6.9. The foregoing derivation 
applies to the tetrode if g m3 is equated to zero or is omitted. 

8 By common usage, v B in a pentode or tetrode tube is understood to be v al . Hence, 
the second subscript may be omitted when no confusion will result. 



Multlelectrode Tubes and Transistors 



209 



Gl 



Vg\ 



?nt«Vl(|) 



Gl 



Vg\ 



/Mlfci 



K 
Current source 



K 
Voltage source 



Fig. 6.9. Equivalent circuits for a pentode or tetrode. 

PROB. 63. Draw the equivalent circuit of a pentode when the voltages of all 
the grids are permitted to vary. 

Example 6.1. Find the small signal voltage gain of the circuit shown in Fig. 6.7. 
The equivalent circuit for the amplifier of Fig. 6.7 is shown in Fig. 6.10. The 
voltage v is found by writing the nodal equation for the plate circuit. 



or 



**«-—(7 m - + Jl) 






(6.7) 
(6.8) 



"»i 



The gain of the pentode amplifier is, therefore, equal to g m times the parallel 
resistance combination of r v and R L . If the pentode amplifier is coupled to 
another vacuum tube through a coupling capacitor C c and a grid resistor R a , 
the reference gain of the pentode amplifier is equal to g m times the parallel 
resistance combination of R L , r v , and R g . (This statement assumes that X ee 
is much less than R„.) 

The usual value of r„ is very large. For instance, the plate resistance of the 
6AU6 is approximately 1 megohm. If the load resistance is quite low in com- 
parison to r„, the parallel combination of r„ and R L is approximately equal to 
R L . Therefore, 

K v « -g m R L (6.9) 




Fig. 6.10. The equivalent circuit of a pentode amplifier. 



210 Electronic Engineering 

PROB. 6.4. Find r v ,g m , and fi for the tube of Fig. 6.4 at a plate voltage of 120 v, 
screen voltage of 75 v, and V G1 = — 3 v. The difficulty encountered in finding n 
indicates why g m is usually used instead of n when working with tetrode and 
pentode tubes. Answer: r„ = 300 K£l,g m - 800 pmho, fi = 240. 
PROB. 6.5. (a) Draw the equivalent circuit for the tube of Fig. 6.4 at V p = 120 v, 
V ai = -3 v, and V G2 = 75 v. Assume R L is equal to 50,000 Q. (b) Find the 
a-c output signal if v G = —3 +0.1 sin cat. 

PROB. 6.6. Determine n,g m , and r v for the pentode whose characteristic curves 
are given in Fig. 6.8. Find values at the point V G2 = 150 v, V G1 = — 2 v, 
V P = 200 v. Answer: r P = 1 MO, g m = 4200 fimho, n = 4200. 
PROB. 6.7. \ig m of a 6AU6 (at a given operating point) is 5200 micromhos and 
r v is 1 megohm, find the voltage amplification of a 6AU6 with a load resistance 
of 20,000 fl. (a) Use Eq. 6.8. (b) Use Eq. 6.9. (c) Compare results. Answer: 
(a) K v = -102 (b) K v = -104 (c) 2% difference. 

PROB. 6.8. Repeat Prob. 6.7 if R L is changed to 500,000 Q. (Note: As indi- 
cated previously, g m is a function of plate current. Hence, as R L is increased, 
g m tends to decrease. Ignore this change of g m in this problem.) 

6.4 GRAPHICAL SOLUTION OF PENTODE CIRCUITS 

The graphical solution of a pentode circuit (or tetrode circuit) is very 
similar to the graphical solution of the triode circuit. However, some 
precautions must be observed when using pentode or tetrode tubes. The 
reason for the precautions can be seen most readily from the dynamic 
transfer characteristic. The dynamic transfer characteristics for load 
resistances of 20,000 and 100,000 ii are shown in Fig. 6.11. (A plate 
supply voltage of 300 v was assumed for both curves of Fig. 6.1 1.) 

The dynamic transfer characteristic for R L = 20,000 Q follows the static 
transfer characteristic very closely except when v G approaches zero. 
However, as R L is increased (to obtain greater voltage gain), the shape of 
the dynamic transfer characteristic changes quite radically. A very little 
change (0.3 ma) of plate current occurs as the grid voltage is reduced from 
to —3 v. From — 3 v to cutoff the dynamic transfer characteristic closely 
follows the static characteristic curve, as shown. The action of the pentode 
for large load resistors can be visualized by referring to the characteristic 
curves. For a load line of R L = 100 K, the operation for v G = to 
v a = —2.5 v is at a point of constant plate current and constant plate 
voltage on the plate characteristics. This point corresponds to the point 
known as collector saturation in the transistor amplifier. 

The voltage gain of the 6AU6 of Fig. 6.11 can be found directly by 
plotting v P vs y f; from Fig. 6.11ft. The resulting curves, shown in Fig. 6.12, 
indicate that the total plate voltage change is essentially a constant for 
R L = 20,000 Q or more. When R L is large, the given change of plate 
voltage requires only a small change of grid voltage. This required grid 



Multielectrode Tubes and Transistors 

20 



211 



8. 



18 

!16 

14 



= 12 

E 

£10 

"c 
« 8 

1 6 

to 

a: 4 
2 



°-5 



























Stalic naiibiei 

characteristic 




































R L = 20K^ 





































-4 



-3 -2 

Grid voltage in volts 

(a) 



-1 








100 



400 



500 



200 300 

Plate voltage in volts 

(b) 
Fig. 6.11. Two dynamic transfer characteristics for a 6AU6 pentode tube, (a) Dynamic 
curve: (6) characteristic curve. (Courtesy of Radio Corporation of America.) 

voltage change is approximately 1 v, peak-to-peak, and is centered approxi- 
mately at V a = —4 v. Therefore, this amplifier would operate very 
satisfactorily, providing the grid bias is held at about —4 v. However, it 
may be seen from Fig. 6.12 that a 0.6 v (15%) change in bias voltage will 
cause the quiescent operating point to move to either extremity of the 
useful operating range. Therefore, difficulty may be experienced in 
establishing and maintaining the quiescent operating point at the desired 



212 



Electronic Engineering 



300 



250 



a 200 
o 



'150 



I 100 



50 



-5 



\ 










\ 




\Rl 


= 20K 










\ 






\ L 


= 100 K 


\ 










\ 


k 










\ 



-3 -2 -1 

Grid voltage in volts 



Fig. 6.12. A plot of v P versus v g for a 6AU6. 



(2.5) 20 |t ^*x-15K(120IQ 
(2.0) 16 



E 
(15) 1 12 

'e 



(1.0) £ 8 



(0.5) 



(0) 





v~>- 


50 v 


V m = 0v 














V H = 0v 


(0) 


















^ \ 








— 0.5 V (— u.i^a;- 




X 








(-0.25)" 






V 






-l.Ov 








\ 




-1.5 v 










\ 


^ 


-2.0 v 


(-0.5) 










\ 


-2.5 v 












1 N 


>3.0v 


(-0.75)" 














- (-1.0) 






















\ 





50 



100 150 200 250 

Plate voltage in volts 



300 



350 



Fig. 6.13. The effect of reducing the screen grid potential on a 6AU6. Open values 
are for v m = 150 v and values in parentheses are for v B1 = 37.5 v. 



Multielectrode Tubes and Transistors 213 

location. In Fig. 6.12, the slope of the curves is the voltage amplification 
of the circuit. 

In many applications it is desirable to operate a pentode with a large 
value of plate load resistance and still use the full grid swing, thus avoiding 
the critical biasing problem. Such operation is possible if the screen grid 
voltage is reduced. The effect of reducing the screen grid voltage is shown 
in Fig. 6.13. 

Example 6.2. In Fig. 6.13, a 15 Kii load line was drawn on a set of 6AU6 
collector characteristics. The original screen grid voltage for this set of curves 
was 150 v and the load line was drawn through the knee of the Vqx = curve. 
The plate current scale was then altered so that the load line previously drawn 
represents a 120 KQ load instead of a 15 KCl load. This required that the plate 
current values be reduced by a factor of 8. The reduced values are shown in 
parenthesis. This reduction in plate current can be obtained by a reduction in 
screen grid voltage. In order to determine the required value of screen grid 
voltage for the modified curves, it was assumed that the plate current is propor- 
tional to the | power of the screen grid voltage. Then 150/(K G2 ) = 8^ = 4, 
and the new value of screen grid voltage (V G2 = ^ = 37.5 v. The values of 
control grid voltage also need to be revised because the grid voltage required for 
plate current cutoff is proportional to the screen grid voltage. Therefore, a 
reduction of screen-grid voltage by a factor of 4 also reduces the cutoff control 
grid voltage by a factor of 4. To meet this requirement, therefore, all values of 
control grid voltage must be reduced by a factor of 4 as shown in Fig. 6.13. The 
.new values of V Gl are enclosed by parenthesis. 

PROB. 6.9. Add the dynamic transfer characteristic of the 6AU6 with V PP = 
300 v, V oi = 37.5 v, and R L = 100 Kii to those presented in Fig. 6.11a. 
PROB. 6.10. Determine proper bias, maximum input signal and maximum out- 
put signal for linear operation of a 6AU6 if R L is 15,000 CI, V PP = 300 v. 
V a2 = 150 v. Answer: V a = -2v, v t <* ±2v, v = 60 to 285v. 
PROB. 6.11. Using the ideas developed in Example 6.2, find a new set of 
coordinates and determine the screen voltage and grid voltages if the 15 Kii load 
line is to represent a 90 KO load line. 

6.5 OPERATION OF A PENTODE FROM A SINGLE VOLTAGE 
SOURCE 

The pentode amplifier circuit shown in Fig. 6.8 requires three batteries 
for proper operation. Through the use of resistors and capacitors, the 
number of batteries or power supplies required can be reduced to one. 
The circuit which accomplishes this saving of batteries is shown in Fig. 
6.14. 

The resistor R K causes a voltage drop due to the flow of cathode current 
and provides self-bias in the manner discussed for the triode tube. 



214 Electronic Engineering 

However, the cathode current for the pentode is 

I K = I P + I a2 (6.10) 

where I K is the quiescent cathode current, I P is the quiescent plate current 
and I G 2 is the quiescent screen grid current. Again, 



R K = 



Fblas 



(6.11) 



Also, C K can be found as shown in Eq. 5.75. However, since r B » R L 
for a pentode, Z t c± \jg m . Thus, 

C = WSmRK + 1) ( 

w iR K 
where w 1 is the lowest frequency the amplifier is designed to pass without 
loss of gain. 




Fig. 6.14. Operation of a pentode from a single voltage source. 

The screen grid voltage dropping resistor R sg causes a voltage drop due 
to the screen grid current flowing through R sg . The value of R sg is found 
by the relationship 

V PP - V a2 

-^ (6.13) 



R,*\= 



where V PP is the value of the plate supply voltage and V a2 is the desired 
potential on the screen grid. Again, I a2 is the quiescent screen grid current. 
In order for the screen grid to have a constant potential,The a-c imped- 
ance to ground in the screen circuit must be low. Otherwise, the a-c 
component of the screen current will produce an a-c component of voltage 
on the screen. Accordingly, the capacitor C sg bypasses the screen grid to 
ground or cathode. However, to find the proper size for C sg , the total 
resistance this capacitor must bypass needs to be determined. 



Multielectrode Tubes ami Transistors 



215 



Since the cathode, control grid, and screen grid have the same configura- 
tion in a pentode as the cathode, control grid, and plate have in a triode, 
an equivalent circuit can be derived for the screen grid circuit. This 
equivalent circuit is shown in Fig. 6.15, provided that v P and v G3 are 
constant. 

The parameter g m2 is given by the relation 



Sm2 — 



dv 



G2 



(6.14) 



ai 



where i Q2 is screen grid current and v G1 is control grid voltage. The 
parameter r g2 is known as the screen resistance and is given by the relation 

dv a2 



r„, — 



diQ2 



(6.15) 









Tube 






o 


[— oGi 










»*1 




c 


^gmWgl * 


f g2 \ 


Z Ctg < 








J 

















Fig. 6.15. Equivalent circuit for the screen grid of a pentode with v P and v a3 constant. 

where v G2 is the screen grid voltage. Figure 6.15 shows that the screen 
grid bypass capacitor C sg is actually in parallel with both R sg and r g2 . The 
rule of thumb, therefore, requires that 



y _ 

^■csg 



Rsg r g2 



lO(R sg + r g2 ) 



(6.16) 



where X csg is the reactance of the capacitor C sg at the lowest frequency the 
amplifier is designed to pass. 

The value of r g2 is seldom included in tube manuals. However, a quick 
experimental check will determine r g2 . In most cases, an estimation of r g2 
is accurate enough for quick calculations. The similarity of the screen 
circuit to the plate of the triode has already been noted. It is not surprising, 
therefore, to find that r g2 has the same order of magnitude as the r p of the 
triode tubes. However, since the screen only intercepts a portion of the 
total space current, the value of r g2 is higher than the value of r p in a typical 



216 



Electronic Engineering 



triode. Usual values of r gi are in the order of 20,000 to 40,000 Q. Of 
course, as the size of the capacitor C M is increased, the screen grid potential 
is held more constant. Therefore, Eq. 6.16 determines the minimum value 
C sg should have whereas cost or size will dictate the maximum practical 
value in a given circuit. 

The quiescent value of screen grid current is given in some tube manuals 
for typical voltages on the plate and various grids. In other manuals, sets 
of characteristic curves include the screen current characteristics with the 
plate characteristics. Such a set of curves is reproduced in Fig. 6.16. 



18 
16 


v« 


= 


150 


V 


V« 


= 


f 




1 


/gi = 0v 




















\ f 












-0.5 v 




|12 

.2 

I 10 

■£ 8 




























-l.Ov 




\ f > 


> 


^ 










-1.5v 




at 

o 6 

•S 

K. 4 


JV^ 






















^ 


-] 


Lv 




-2.5 v 


~ -2.0 V 














-3.0v 






2 


._ 


3v 
















K """ 


















U 






1 






-4.0 v 






D 







1C 


10 


1! 


50 2( 
Plate volt 


X) 

age in 


25 
volt 


3( 


X) 3 


50 


40 



Fig. 6.16. Characteristic curves of a 6AU6 pentode including grid current charac- 
teristics. 

The method outlined for operating a pentode from a single power supply 
can, of course, be used for operating a tetrode or any multigrid type of 
tube from a single power supply. 

PROB. 6.12. Determine values for all the resistors and capacitors in the 6AU6 
circuit of Fig. 6.14 if V PP is 250 v and R L is 25,000 O. (Bias the tube for maxi- 
mum linearity.) Assume r g2 — 30 KQ. Find the voltage gain and maximum 
power output. The lowest frequency to be amplified is 30 Hz. Answer: R K = 
380 O, C K = 140 (if, R sg = 45.5 KO, C sg = 3 /if, G v = 63. 

6.6 BEAM POWER TUBES 



Beam power tubes are large current tetrodes which are constructed so 
the screen grid wires have the same spacing as the control grid wires. The 



Multielectrode Tubes and Transistors 



217 



screen grid is then oriented so its wires are in the electrical "shadow" of 
the control grid wires. The negative potential of the control grid repels 
the electrons and forms an electron beam in the space between the control 
grid wires. Since the screen grid wires are directly behind the control grid 
wires, the screen grid intercepts a minimum number of electrons. The 
screen grid also tends to diffuse the electron beam somewhat. The relative 
locations of the grid wires and the general shape of the electron beams are 



-Beam- forming plate 




.17. The action of the beam power tube. 



shown in Fig. 6.17. To enhance the beam-forming qualities, beam- 
forming plates are also included as shown in Fig. 6.17. 

If the plate potential of the beam power tube is quite low, the speed of 
the electrons decreases as the electrons approach the plate. The charge 
density, as noted in Chapter 3, is given by Eq. 3.51. 



= —P 



(3.51) 



Therefore, as the velocity of the electrons is decreased, the charge density 
is increased. Consequently, when the plate voltage is low, a negative 
charge is produced in the space between the plate and screen grid. If the 
screen grid is more positive than the plate, we might expect the electrons to 
reach their minimum velocity at the plate. However, electrons leaving the 



218 



Electronic Engineering 



screen grid area experience a decelerating force due not only to the lower 
potential of the plate but also due to the negative charges on all the 
electrons between the screen grid and plate. Consequently, a minimum 
velocity is achieved before the electrons reach the plate. In fact, electrons 
are actually accelerated from this minimum value as they approach the 
more positive plate. Therefore, minimum potential in the screen-to-plate 
area occurs between the screen grid and the plate. 

This space charge has the same effect in the beam power tube as the 
suppressor grid exerts in the pentode tube. The secondary electrons are 

400 



300 



200 



£ 100- 



UOl - 


+ 15 














-±^1 


_ + 5 

lo 1 volts o ( 


;i=° 
















li 


10 












-14;*. 






-lb 










u G1 = -20 








r*- 28 


F ^ 


-25 
-30- 



100 



200 



300 400 

Plate voltage in volts 



500 



600 



700 



Fig. 6.18. Characteristic curves of a 6L6 beam power tube. (Courtesy of Radio Cor- 
poration of America.) 



returned by the space charge to the more positive plate. As a result, the 
characteristic curves of a beam power tube are quite similar to the charac- 
teristic curves of a pentode tube. Characteristic curves of a beam power 
tube are shown in Fig. 6.18. Actually, the fairly linear portion of the plate 
characteristics of a beam power tube extend to lower plate voltages than 
the comparable pentode power tube. Therefore, greater power output and 
higher efficiency can be obtained from the beam power tube than can be 
obtained for a conventional pentode having similar ratings. 

For proper suppression of secondary emission the beam power tube 
requires a large plate current compared to a pentode which might be used 
for voltage amplification; consequently, the word "power" in the title of 
the tube. Therefore, the pentode is usually used as a voltage amplifier 
whereas beam power tubes are usually used in output circuits which require 
a relatively large amount of power. 



Multielectrode Tubes and Transistors 219 

PROB. 6.13. Design a beam power stage similar to the pentode stage of Fig. 
6.14 for a 6V6 tube if V PP is 300 v and R^ is 3000 Q, and J G2 = 4.5 ma. (a) 
Determine values of all resistors and capacitors used, (b) Determine the maxi- 
mum output voltage signal, (c) Find voltage gain, (d) Find maximum signal 
power output. Assume r g2 = 30 KC1. The lowest frequency to be amplified is 
30 Hz. 

6.7 TUBES WITH MORE THAN THREE GRIDS 

Special purpose tubes are constructed with more than three grids. In 
fact, pentagrid converter tubes are made which have five grids in addition 
to the cathode and plate. These tubes are designed for use in the first 
detector stage of superheterodyne receivers. The primary use is to mix 
two or more electrical signals to produce new signals. Typical uses of these 
tubes will be covered in more detail in the later chapters in this book. 

Many of the modern tubes contain several sections. For example, two 
triodes in a single envelope is very common. In fact, a 6SN7 is two 6J5 
tubes in a single envelope. These two sections operate as two separate 
tubes as far as the electrical circuit is concerned. However, in terms of 
economy, one 6SN7 requires only one filament and one tube socket com- 
pared to two filaments and two tube sockets for the two 6J5 tubes. Also, 
the cost of two 6J5 tubes is higher than the cost of one 6SN7 tube. In 
addition, the space required for one 6SN7 is about one-half the space 
required for two 6J5 tubes. Therefore, multisection tubes are more 
economical and require less space than the several single-section tubes 
which they replace. 

6.8 FIELD-EFFECT TRANSISTORS 

The field-effect transistor (FET) is a voltage controlled semi-conductor 
which has a high input impedance, comparable to that of a vacuum tube. 
This transistor has only one p-n junction as shown in Fig. 6.19, and 
therefore, it is frequently known as a unipolar field-effect transistor 
(UNIFET). 



Gate 



Source 




Fig. 6.19. Typical field-effect transistor structure. 



220 Electronic Engineering 

The schematic representation in Fig. 6.20 will be used to explain the 
principles of operation of the field-effect transistor. Figure 6.20a shows 
that a narrow semiconductor channel provides a conducting path between 
the source and the drain. This channel may be either an n- or a p-type 
crystal. The n-type is used in this discussion. With no biases applied 
to the transistor, the channel conductance G e = a(wt/l), where a is the 
conductivity of the crystal and w, t, and / are the width, thickness, and 
length of the channel, respectively. If a reverse bias is applied between 



L 



Gate 

1 



<-- , J - 



Source 



Gate 



Source 




1 > * I ' 

(a) v GS = 0, v DS = , 



Drain 1 



Channel 
Drain 



(b) v GS = V^> (Pinch off), v DS = 




C 



(c) v GS = 0, v DS = V P 





(d) v GS = 0,v BS >V P (e) vgg <0, v DS > V P 

Fig. 6.20. Schematic representation of the field-effect transistor. 



the gate and the source, the depletion region width is increased and the 
thickness of the channel is decreased. The reader . will recall that the 
excess carriers (electrons for «-material) have been removed or "depleted" 
from the depletion region. Therefore the depleted region will not con- 
tribute to the conduction. The gate bias required just to reduce the channel 
thickness to zero is called the pinch-off voltage V P (Fig. 6.206). 

If the gate-source voltage v GS is zero and the drain is made positive 
with respect to the source, electrons drift through the channel because of 
the electric field. The drain current I D is equal to the drain source voltage 
v DS times the channel conductance G c , providing that v GS is very small. 
However, the positive drain voltage reverse biases the p-n junction near 
the drain end of the channel, and when the drain voltage is increased to 



Multielectrode Tubes and Transistors 221 

the pinch-off voltage the channel thickness is reduced to zero at a point 
near the drain end of the channel. The drain current does not stop when 
the drain voltage reaches pinch-off because a voltage equal to V P still 
exists between the pinch-off point and the source. Therefore, carriers are 
injected from the source into the depletion region. The resulting electric 
field continues to accelerate the carriers through the channel to the drain 
(Fig. 6.20c). As the drain voltage is increased beyond V P , the depletion 
region thickness is increased between the gate and the drain, but is prac- 
tically unchanged between the pinch-off point and the source (Fig. 6.20d). 
Therefore, the source current remains essentially constant as the drain 
voltage increases above V P because the additional voltage appears across 
the depletion region and the electric field along the channel does not 
change. This saturated drain current is known as I DSS (saturated short- 
circuit drain current), which is measured at v DS = V P , v GS = 0, and 
* t = 0. 

The field-effect transistor normally oper ates with the drain voltage v n s 
ab ove V P and reverse bias applied to tne gate. The electric field (and 
thus the drain current in the channel) is then controlled by the gate-source 
voltage v GS in a manner similar to the control of the plate current in a 
vacuum tube by the electric field between the grid and the cathode. The 
effect of the gate voltage v GS on the channel conductivity is shown in 
Fig. 6.2(te. The drain current is almost independent of drain voltage 
whenever the sum of the drain voltage and the reverse-bias gate voltage 
exceeds V P . 

The drain characteristics of a typical n-channel transistor are shown in 
Fig. 6.21a. Notice the similarity of these characteristics to the pentode 
characteristics. Like the pentode, the output resistance r d is very high in 
the normal operating range of the transistor. Avalanche breakdown 
occurs at the junction whenever the drain-gate voltage exceeds a given 
value (about 35 v for the transistor of Fig. 6.21). The transfer character- 
istics are shown in Fig. 6.216. Observe that, like the pentode tube, a 
single curve is adequate to describe the transfer characteristics for all 
drain voltages in the normal operating region. The slope of the transfer 
curve at any point is the trans-conductance g m of the transistor at that 
point. 

The input characteristics of the typical n-channel FET are shown in 
Fig. 6.21c. The gate current is the reverse-bias saturation current of the 
junction. This current is of the order of 10~ g amp in a small-signal 
silicon transistor. The input conductance is the slope of the input charac- 
teristics, which is very low in the normal operating range. The input 
resistance r a is the order of 10 8 to 10 10 ohms. N ote that the input resistance 
remains high when forward mas not exceeding a few tenths of a volt is 



222 




Electronic Engineering 



10 20 30 

Drain volts , , 

(a) V/D-- 



40 




J_ 



-ii- 



J L 



"7^30 "-1 -0.5 
I Gate volts 

(c) 



0.5 



Fig. 6.21. Characteristics of a typical n-channel field-effect transistor, (a) Drain 
characteristics, (b) Transfer characteristics, (c) Input characteristics. 



applied to the gate. Therefore, v GS = is a suitable Q point for a small- 
signal amplifier. Note that the input resistance drops abruptly when 
avalanche breakdown occurs. Avalanche breakdown occurs whenever the 
difference between the gate and drain potentials exceeds the avalanche 
breakdown voltage. 

A d-c conducting path between the gate and the source must be provided 
in order to maintain the desired bias. The drain-gate saturation current 



Multielectrode Tubes and Transistors 



223 



Gate 



Drain 




Source 



Fig. 6.22. y-parameter equivalent circuit of an FET. 

I DO must flow through this path; therefore, the external gate-source 
resistance should not exceed a few megohms. 

The equivalent circuit of Fig. 6.22 is a general y-parameter circuit 
which is applicable to the field effect transistor providing that the signal 
levels are small and, therefore, the y-parameters are linear. At low 
frequencies the depletion region capacitances may be neglected and the 
equivalent circuit is simplified (Fig. 6.23a). The high-frequency limitations 
of this transistor will be discussed in Chapter 7. Note the similarity of 
both the characteristic curves and the equivalent circuit of the field effect 
transistor as compared with the pentode tube. The output conductance 
g a and the transconductance g m can be determined from the curves given 
in Fig. 6.21 by the same techniques described for the vacuum tubes. The 
gate conductance g 9 is so small over the normal range of operation that 
it cannot be determined from Fig. 6.21c. 

A common source circuit is shown in Fig. 6.236. In this circuit, the 
gate bias is zero, which is satisfactory for very small input signals because, 
as shown in Fig. 6.21c, the gate conductance g g is negligible for forward 
bias voltages below about 0.2 v in a silicon transistor. A load line can 




o 



.R L 



Re'. 



Source 

(a) m 

Fig. 6.23. (a) Equivalent circuit, (b) Actual circuit for a field-effect transistor. 



224 



Electronic Engineering 



be drawn on the drain characteristics and the voltage gain can be deter- 
mined graphically, as previously illustrated for the vacuum tube. Also, 
the voltage gain can be determined by the use of the equivalent circuit, 
Fig. 6.23a. 



r d R 



a^L 



(6.17) 



(6.18) 



v. = g m v r 

r d + R L 

where r d = l/g d . 
Normally, r d is very large in comparison with R L ; therefore, 

r »o g m r d R L „ „ „ 
G v = - = — — — ~ g m R L 

v< r d + R L 

Equation 6.18 holds only when the minimum drain voltage exceeds V P . 
Otherwise, r d is not large in comparison with R L . 

Figure 6.24a illustrates a technique for selecting a suitable transistor 
and a satisfactory operating point for a specified load resistance. A wide 
range of drain saturation current (I DSS ) values is available in the various 




Drain voltage excursion - 
(a) 




Fig. 6.24. (a) Load line on drain characteristics, (b) Typical circuit using source bias. 



Multielectrode Tubes and Transistors 225 

FET types. Therefore, a transistor which will permit the specified load 
line to pass near the knee of the v os = curve (v as = 0.5 curve for 
silicon FET's) is usually available. After the load line is drawn, the Q 
point is selected so that the entire excursion of drain voltage will remain 
in the relatively flat portion of the drain characteristics (Fig. 6.24a). 

Source bias (like cathode bias) may be utilized as shown in Fig. 6.246. 
The bias resistance may be determined in the same manner as for a vac- 
uum tube. 

Rs = ^ (619) 

The bypass capacitor must bypass the impedance looking into the transis- 
tor source terminal in parallel with R s . This source impedance is Ao os / 
Af s c=l Av os /Ai D ~ l/g m . Therefore, the value of bypass capacitance. C s 
which will provide a low frequency cutoff co t is 

C = (g " + G * )10 (6.20) 

where G s = l/Rs- 

The variation of drain current with temperature in a field effect transistor 
is determined by two factors. One factor is the temperature variation of 
depletion region width, which results from the temperature variation of 
barrier height (V^ — V), discussed in Chapter 3, where V^ is the zero- 
bias barrier height. As previously discussed, the temperature coefficient 
of this voltage is about —2.2 mv/°C, which results in an increased drain 
current with increased temperature. The other factor is the variation of 
majority carrier mobility with temperature. This mobility influences the 
transconductance g m . As the temperature increases, the carrier mobility, 
and hence g m , decrease and tend to compensate for the variation of V^ 
with temperature. In fact, the proper choice of Q point will give essentially 
zero temperature coefficient of drain current from about —50 to 100°C. 
The temperature coefficient due to mobility change is about 0.7%/°C. 
Therefore, the condition for zero temperature coefficient is 

0.007(-i D )/°C = <?m (0.0022)/°C (6.21) 

l -2- = -0.315 v (6.22) 

Sm 

The relationship of Eq. 6.22 can be expressed more conveniently in 
terms of the drain saturation current I DS s and the transconductance g mo . 
A typical relationship of i D to I DSS for a diffused gate transistor is 



'-('-Si 



-'M.U-™ («-23) 



226 



Electronic Engineering 



Source Gate Drain 




Depletion 
region 



-10 



-8 



E 

■£ -6 



-4 







-12v 














-lOv 
















-8v 










-6v 








Vgs = 


_4v 

















-2 



-10 -20 -30 -40 -50 
Drain voltage-volts 

(b) 
Fig. 6.25. (a) MOS schematic construction, (b) Typical drain characteristics. 

where T DSS * s tne value of i D at v GS = 0. Also, the transconductance g„ 
is related to v GS by the following equation ! 



&m omol A t/ / 



(6.24) 



where g mo is the value of g m at v GS = 0. Then, substituting Eqs. 6.23 
and 6.24 into Eq. 6.22, 



lms( l _Eos) = o.315v 



But, using Eq. 6.23, 



omo 



di 7 



-21 



nss 



dv GS \vos=o V P 

Substituting this value of g m0 into Eq. 6.25, 

V P — v ns = —0.63 v 



J as 



(6.25) 
(6.26) 

(6.27) 



Multielectrode Tubes and Transistors 227 

or 

vos = v i> + °- 63 v ( 6 - 28 ) 

Equation 6.28 shows that zero thermal drift may be achieved if the FET 
is biased 0.63 v above the pinch-off voltage V P . Note that V P and v GS 
are negative for an w-channel FET. All signs should be reversed for a 
/^-channel FET. 

A special type of field effect transistor is the metal-oxide-semiconductor 
(MOS) (Fig. 6.25). In this type, the depletion region is replaced by a thin 
layer of silicon oxide, which is a good insulator, and the semiconductor 
which formed the gate is replaced by a metal conductor. A schematic 
representation of a typical structure is shown in Fig. 6.25a. The principle 
of operation is similar to a conventional field effect transistor but is 
somewhat more complex. A detailed description of the principle of 
operation may be found in the literature. 4 The insulating layer permits 
large-signal operation with forward bias, provides extremely high input 
resistance, of the order of 10 1S ohms, and reduces the input capacitance 
of the device. A typical set of drain characteristics for a MOS field-effect 
transistor is shown in Fig. 6.25b. 

PROB. 6.14. The transistor of Fig. 6.21 is used as an amplifier with a resistive 
load and V nn = 35 v. Design the amplifier for the Q point v as = 0. Draw 
a circuit diagram and determine the voltage and power gains. Answer: R L = 

6.8 KCl, G v = -12.3. 

PROB. 6.15. The amplifier of Prob. 6.14 is to be biased for zero drift. Select 
the Q point, draw a circuit diagram, and determine the values of all components. 
Use source (like cathode) bias and bypass for frequencies down to 30 Hz. 
Determine the voltage and power gains. 

6.9 THE UNIJUNCTION TRANSISTOR 

The unijunction transistor is composed of a bar of «-type semi-conductor 
material with an electrical connection on each end. The leads to these 
two connections are called base leads — base-one B x and base-two B 2 . A 
third connection is made at a position between the two end connections 
where a single p-n junction has been formed. The lead to this junction 
is called the emitter lead E. The construction of the unijunction transistor 
is reflected in the symbol which is given in Fig. 6.26a. The equivalent 
circuit for the transistor is given in Fig. 6.26b. 

Under normal conditions, the base-one B y is connected as the common 
terminal. Then, if the emitter is opened, the transistor acts as a voltage 

4 Electronics, McGraw-Hill, November 1964. 



228 



Electronic Engineering 




-0B2 



.Rm 



=- Vbb Eo W" 



P- 



-0B1 



(a) 



(b) 



Fig. 6.26. (a) The symbol and biasing arrangement of a unijunction transistor. (6) An 
equivalent circuit for the unijunction transistor. 

divider network and the emitter voltage is a given fraction of the base- 
two voltage. 

Veo = VVbz (6-29) 

If the emitter is biased at a potential less than V EO , the p-n junction is 
reverse biased and only the diode saturation current flows in the emitter 
circuit. If the voltage of the emitter is increased above V E0 , the junction 
becomes forward biased. Under these conditions, holes are injected from 
the ^-material into the n-bar. These holes are repelled by the positive 
base-two end of the bar and they are attracted toward the base-one end 



4000 
i 3000 

JZ 



c 

55 2000 

1000 









































































































10 20 30 40 

Emitter current in milliamperes 

Fig. 6.27. A plot of R Bl vs emitter current for a 2N492. 



50 



Multielectrode Tubes and Transistors 



229 



Saturation 
current 



Yeo 


„^-Peak point 1 




\ Negative 1 
■«— \resistance*+< — Saturation region — »- 
\ region j 




\ j /Valley region 


n— - 


Emitter current *- 



Fig. 6.28. The emitter current-voltage characteristics of a unijunction transistor. 

of the bar. This accumulation of /^-carriers in the emitter-to-base-one 
region results in a decrease of resistance for resistor R m (Fig. 6.27). 
The decrease of resistance R m results in a lower emitter voltage. Thus, 
a negative resistance results since more emitter current results in lower 
voltage. As more /^-carriers are injected, a condition of saturation will 
eventually be reached. This action is shown in Fig. 6.27 and in the 
current-voltage characteristics in Fig. 6.28. The similarity of these 
characteristics to those of the tunnel diode is at once evident. The uni- 
junction transistor, however, has three terminals. Therefore, in order to 
completely describe the characteristics, a family of curves (as given in 
Fig. 6.29) is required. 



20 



16 



-12 



Z 8 











































\ 




















Vi 


V BB = 30 v 
















\\ 


-V BB = 20 v 
















V 




















N 






= 5v 
















\<^ 


><TJb2 = 






























<^~ 



















6 8 10 12 14 16 

Emitter current in milliamperes 



18 20 



Fig. 6.29. Static characteristic curves of a unijunction transistor. 



230 



Electronic Engineering 




Fig. 6.30. A relaxation oscillator. 

Unijunction transistors are used extensively in oscillator, pulse, and 
voltage sensing circuits. A typical application is illustrated by the relaxa- 
tion oscillator of Fig. 6.30. This circuit will be used to illustrate a typical 
use for a unijunction transistor. 

Example 6.3. A transistor is connected as shown in Fig. 6.30. If V SB = 12 v, 
Hi = 100 Kii, R 2 = 2.5 Kii, and C = 0.01 /ufd, plot the output voltage across 
capacitor C. The transistor is a 2N492 with the following characteristics: 
Rm + Rb* = Rbb ~ 7.5 KQ, v <* 0.67. 

At time t = 0, the 12 v battery is applied. The capacitor is uncharged and a 
current of 12/(7.5 YSi + 2.5 Kfl) = 1.2 ma flows through R 2 . The base B 2 is at a 
potential of 9 v and V EO (from Eq. 6.29) is 9 x 0.67 = 6 v. Thus, the emitter 
junction is reverse biased and can be considered as if it were an open circuit. 
The capacitor C charges through/?! with a time constant of /J t C = 10 5 x 10~ 8 = 
10 -3 = 1 msec. Since C is charging toward 12 v, the equation for v c is 



v c = 12(1 - e 



-«/io- 



'>) 



(6.30) 



■100K 



o.oiMfd-r-ev 



\R 2 +Rm = 5K 



■ R m = 40a 



+ + 

5-12v 6v=;0.0lMfd 




0.095v 



(b) 



Fig. 6.31. (a) The equivalent circuit for the emitter of the transistor in Fig. 6.30 when 
conduction occurs. (6) A simplified form of this equivalent circuit. 



Multielectrode Tubes and Transistors 



231 



The voltage on the capacitor will be 6 v in 692 ix seconds. At this time, t lt the 
emitter becomes forward biased and current begins to flow into the emitter. 
The resistor R Bl decreases and more current flows into the emitter. This action 
is accumulative and occurs very rapidly driving the resistor R m into saturation. 
From Fig. 6.27, the saturation resistance is about 40 ohms or so. Thus, the 
emitter circuit becomes approximately that shown in Fig. 6.31a. When a The- 
venin's equivalent circuit is constructed for the network that connects to the 
capacitor, the circuit is simplified to the form shown in Fig. 6.316. This circuit 
has a time constant of 4 x 10 x 10 -8 = 0.4 ^seconds. The capacitor is dis- 
charging toward essentially zero potential. Thus, in about 5 time constants, or 
2 n seconds, the capacitor C is essentially discharged. The excess holes are 
swept out of the emitter-to-base-one region and the resistance Rbi returns to its 




694 1388 

Time in microseconds 

Fig. 6.32. The waveform of the circuit shown in Fig. 6.30. 

normal condition. The circuit is now the same as at time t = 0, so the entire 
process is repeated. Thus, the voltage waveform across the capacitor has the 
form shown in Fig. 6.32. 

The relaxation oscillator can be used in conjunction with other wave- 
shaping circuits to produce a variety of pulse circuits. 5 On a note of 
caution, the power ratings of the transistor should not be exceeded. 
Since R m may become very small, the voltage drop across and the current 
through R m may become large enough to damage the transistor. 

PROB. 6.16. (a) Determine the power dissipated in the transistor of Example 
6.3 when the emitter is cutoff. Answer: 10.8 mw. 

(b) Determine the power dissipated when the emitter junction first becomes 
saturated. Answer: 14.4 mw in R B2 and 900 mw in R m . 

(c) If the power in part b is assumed to be dissipated during the entire 2 n 
second discharging time, what is the average power dissipated in this transistor ? 
Answer: 13.4 mw. 

PROB. 6.17. Repeat Example 6.3 if R 2 is changed to 7.5 Kfl, and R x is changed 
to 500 KQ. 



5 For additional unijunction circuits see G. E. Transistor Manual, Seventh Edition, 
General Electric Co., Syracuse, New York, 1964, pp. 301-347. 



) 



132 



Electronic Engineering 



6.10 THE SILICON CONTROLLED RECTIFIER 

A semiconductor device known as a "Thyrode" or Silicon controlled 
rectifier (SCR) is constructed as shown in Fig. 6.33a. The action of this 
device can beexplained by the equivalent circuit which is shown in Fig. 6.33ft. 
This equivalent circuit is composed of an n-p-n transistor and p-n-p transis- 
tor connected as shown. In one condition, (gate negative) the bias to 
the gate (the base of the n-p-n) maintains the n-p-n transistor in the cutoff 



Gate 



y ^2^+ 



Cathode 









Anode + 


















Emitter 
P 






Collector 
re 




Base 
re 










+ Gate 


Base 
P 


Collector 
P 












Emitter 
re 






athode — 















C 



<a) 

Fig. 6.33. The silicon-controlled rectifier, 
electrical circuit. 



(b) 

(a) Actual construction; (b) equivalent 



condition. Since no collector current flows in the n-p-n transistor, the 
p-n-p transistor is also cutoff. Therefore, the impedance between anode 
and cathode is very high. 

If the bias on the gate is changed so the n-p-n unit starts to conduct, a 
collector current will flow into the n-p-n collector from the base of the 
p-n-p unit. This flow of current from the p-n-p base causes collector 
current to flow in the p-n-p transistor. The collector current of the p-n-p 
transistor is fed to the base of the n-p-n transistor. The action is accumu- 
lative since an increase of current in one unit causes an increase of current 
in the other unit. As a result of this action, both transistors are driven 
to current saturation. Consequently, the impedance between anode and 
cathode becomes very low. 

As soon as the self-regeneration commences, the gate (the reason for 
this name is now obvious) loses all control over the action. The collector 
current from the p-n-p unit is much larger than the external gate current. 



Multielectrode Tubes and Transistors 



233 



As a result, the external gate circuit can turn the SCR on but cannot turn 
the switch off. To turn the SCR off the gate bias must be in the reverse 
direction and the anode voltage reduced essentially to zero. In this 
condition, no current flows through the p-n-p unit and the gate regains 
control of the circuit. 

If the gate is maintained at cutoff (negative potential) and the voltage 
on the anode is varied, the characteristics are as indicated in Fig. 6.34. 



_ Avalanche 
'breakdown 



- on characteristics 



Breakover 
- voltage 



E — *- + 

off characteristics 



Fig. 6.34. The curve of / vs V for the silicon-controlled rectifier. (Courtesy of Solid 
State Products, Inc.) 

At a positive anode potential, which corresponds somewhat with the 
avalanche breakdown potential in the reverse bias case, the SCR will turn 
itself on. The potential required is known as the "breakover voltage." 
In effect, the current through the device does start to increase as in the 
regular avalanche breakdown. (Some junctions are forward biased and 
some are reverse biased.) However, this increasing current through the 
p-n-p section produces base current for the n-p-n section, and the regenera- 
tive action quickly takes over turning the SCR on. The maximum voltage 
ratings of the device are chosen below the avalanche breakdown voltage 
and the breakover voltage. 6 Consequently, the SCR can only be turned 
on with the gate as long as recommended maximum voltage ratings are 
not exceeded. An example will illustrate the usefulness of the silicon 
controlled rectifier. 

" A p-n-p-n device with only two leads (corresponding to the anode and cathode) has 
also been developed. This device is known as a Shockley diode and uses the breakover 
voltage to switch the device on. It is also possible to purchase a light-operated p-n-p-n 
device. 



234 



Electronic Engineering 



150sin377t 




35 K 



-WNAAA- 




Gate 



< Anode 

3A200 
6 Cathode 



Fig. 6.35. A silicon-controlled rectifier connected to a load. 

Example 6.4. A 3A200 SCR is connected in a circuit as shown in Fig. 6.35. 
The 3 A200 has the following ratings : 

Peak anode current 1000 ma 

Power dissipation (case 100°C) 2.5 w 

Peak anode voltage ±200 v 

Peak gate current 25 ma 

Peak gate voltage ±5 v 

Maximum gate current to fire +2 ma 

Maximum gate voltage to fire +3 v 



1000 



100 






fl 



10 

1 

0.1 

0.01 

0.001 



: 0.0001 













3 
2 




















e 
f ! i 

g'E 

I 

-1 










i 


















( 


— 























2 3 4 

Anode voltage in volts 
(a) 



-50 50 100 150 

Junction temperature in °C 

(b) 



_ 



g -0.0001 
I -0.001 



E 

■s 

£ 

U 

1 



50 100 150 200 250 
Anode voltage in volts 

(c) 



300 



-0.01 

-0.1 

-1 

-10 



















































1 













-300 -250 -200 -150 -100 -50 
Anode voltage in volts 
(d) 



Fig. 6.36. Electrical characteristics of the 3A200 SCR. (a) Forward on; (b) gate 
current to fire; (0 forward off; and (d) reverse. (Courtesy of Solid State Products, 
Inc.) 



Multielectrode Tubes and Transistors 



235 



The electrical characteristics are as shown in Fig. 6.36. The load resistor R L is 
1000 a. 

Assume the adjustable contact of the 5 Kti resistor is adjusted so the gate is 
effectively connected to the junction of the 5 KC1 and 35 K£i resistors. 

The approximate gate input impedance can be found by dividing the maximum 
gate voltage to fire by the maximum gate current to fire. Hence, 



gate 2 x 10~ 3 



= 1500 n 



(6.31) 



The actual gate circuit can be drawn as shown in Fig. 6.37. The parallel resist- 
ance of the 5 K and 1.5 K resistors is 1.15 K. The voltage across the gate- 
cathode junction is 

150 

1.15 K sin 377r 



or 



35 K + 1.15 K 

4.77 sin 377/ 



(6.32) 
(6.33) 



160 sin 377* 



•35K 



'5K 



.Gate 

', Resistance of 
• the gate 1.5 K 



ToR, 



Cathode 
Fig. 6.37. Equivalent gate circuit for Fig. 6.35. 



Now, from Fig. 6.366, the gate current required to fire the SCR is 0.65 ma when 
the temperature is 50°C. (This 50°C is the assumed operating temperature.) If 
the gate resistance is constant, 7 the voltage across the gate-cathode junction will 
be 

6.5 x 10-* x 1.5 x 10 s = 0.975 v (6.34) 



This value of 0.975 v will be achieved when 

0.975 = 4.77 sin 6 



or 



= 11.8° = 0.206 radians 



(6.35) 
(6.36) 



Therefore, the SCR will be turned on when 377? = 0.206 or t = 0.000547 sec. 
The SCR will remain on until the anode potential is reduced essentially to zero. 

7 The gate resistance will, of course, change with the temperature and with the gate 
current magnitude. However, because operating temperatures are not precisely known, 
the solution is not precise. Hence, the error introduced by assuming r gate = 1.5 Kfi 
can usually be ignored. 



236 



Electronic Engineering 



+ 150 



150 sin 377t =g 

> 









\ t in seconds 




f 


A 









1 1 


/l 


1 3 


11 




120 \ 


/60 


U20 


'30 



/Gate fires above 
this voltage 



Vgate 




Vload 



Fig. 6.38. The voltages and current in the circuit shown in Fig. 6.35. 

(Actually, according to Fig. 6.36a the SCR will turn off when the anode potential 
is less than 0.8 v.) 

When the SCR is turned on, the voltage across the SCR is 0.8 v for a 
current of 3 ma and increases to 1.5 v at 150 ma anode current (Fig. 6.36a). At 
the instant the SCR is turned on, the voltage across the load and SCR is 



150 sin 0.206 = 30.7 v 



(6.37) 



Multielectrode Tubes and Transistors 237 

Approximately 1-v drop exists across the SCR and the remaining 29.7 v are 
across the load R L . The current through the load is equal to 29 Kooo or 
29.7 ma. At the instant peak voltage is applied, the voltage drop across the 
SCR is 1.5 v and the remaining 148.5 v is across the load. The peak current is, 
therefore, equal to 148.5 ma. 
The voltages and current of the circuit are plotted in Fig. 6.38. 

The 5 KD resistor in Fig. 6.35 can be adjusted so the SCR will not be 
turned on until the voltage is almost at the peak value. The average current 
through the load can, therefore, be controlled over about a 2: 1 range by 
the circuit just considered. 

If the voltage to the gate is shifted in phase as well as amplitude, the 
current may be controlled from maximum to almost zero. 

PROB. 6.18. The tap on the 5 KQ resistor of Fig. 6.35 is set 1 Kti above ground. 
Find the voltages and currents of the circuit under these conditions. Plot curves 
similar to those in Fig. 6.38 for these conditions. 

PROB. 6.19. The voltage to the load and switch of Fig. 6.35 is 150 sin 377/. 
The voltage to the gate has been shifted 

V m u = 4.77 sin (377f - 1 radian) 

Find the voltages and currents of the circuit under these conditions. Plot curves 
of voltages and currents as functions of time. 



6.11 THE GAS TRIODE OR THYRATRON 

A gas-filled triode has characteristics quite similar to those of the SCR 
just discussed. As long as the control grid of the gas triode is main- 
tained sufficiently negative, the tube is cut off and no plate current 
flows. If the plate has a positive potential and the control grid bias is 
reduced, a stream of electrons will start to flow between the cathode and 
plate. If the plate potential is well above the ionizing potential of the 
gas, these electrons will cause the gas to ionize and a self-sustaining glow 
discharge will result. The tube is said to "fire" when the glow discharge 
commences. 

Once the gas in the tube has been ionized, positive ions will be attracted 
to the negative control grid. Since these positive ions travel slowly com- 
pared to the electrons, the space around the control grid will contain a net 
positive charge. Asa consequence, the field of the control grid is neutralized 
by this positive space charge. Therefore, the control grid loses control as 
soon as the tube "fires." These positive ions also cause a current flow in 
the grid circuit. As a result, most thyratron circuits have a resistor inserted 
in series with the grid to limit this grid current to a safe value. 



238 




Electronic Engineering 

500 



V 
100 « 



-6 -4 -2 
d-c grid No. 1 supply volts 

Fig. 6.39. Average firing characteristics of a 2D21 gas thyratron. 

Once the tube has fired, the plate must be reduced below the ionizing 
potential of the gas before the control grid can regain control. The actual 
firing potential of the control grid is dependent on the potential on the 
plate. Typical firing conditions for a 2D21 are given in Fig. 6.39. The 
actual firing voltages are dependent on gas temperatures, gas pressures, 
condition on electrodes, etc. 

PROB. 6.20. A gas triode is connected as shown in Fig. 6.40. Assume the tube 
has the same characteristics as the 2D21 of Fig. 6.39. 

(a) Plot the applied voltage as a function of time. 

(b) Plot the current through the 26,500-ii resistor and the capacitor C as a 
function of time if C is (1) 1 nf, (2) 0.1 fif, (3) 0.01 /if. (Assume the grid draws 
no current for this part.) 

(c) Plot the current through the 100- A load resistor for the three conditions 
of part b. When the tube fires, the plate to cathode potential is 10 v. 

(</) Find the magnitude of the resistor R 8 which will limit the maximum grid 
current to 2 ma. Neglect the internal resistance of the tube. 



110 v rms 
60cps 



« t =100Q 
; 26,500 a 



ztzc 



-AAAAr 




Fig. 6.40. The circuit for Prob. 6.20. 



Multielectrode Tubes and Transistors 



239 



PROB. 6.21. Design a pentode amplifier stage which will give a voltage gain of 
100. Use a 6AU6 tube and do not apply more than 300 v to the plate circuit. 
Indicate the size of all resistors and capacitors if 60 Hz is the lowest frequency 
to be used. 
PROB. 6.22. The "typical operating conditions" of a 6J7 pentode are listed as 



Plate voltage = 150 v 
Screen voltage = 100 v 
Grid voltage = — 2 v 



Plate current =3.5 ma 
Screen current =0.5 ma 



(a) Draw the circuit diagram for a pentode amplifier. (6) Use above voltages and 
currents to determine values of all resistors and capacitors in the circuit. A V PP 
supply of 250 v is available and the lowest frequency to be amplified has an co of 
500. Assume At> G2 /A/ G2 = 40,000. 



v in = — 3 + sin uit 




6AU6 



►30 K 



_+ | + 

-=_150v -=-300v 



J- 



Fig. 6.41. The circuit configuration for Prob. 6.23. 



PROB. 6.23. A 6AU6 tube is connected as shown in Fig. 6.41. Sketch the 
output signal listing maximum, minimum, and quiescent values. 



7 



Small-Signal Amplifiers 



Electronic amplifiers are customarily used to amplify the signal from 
a transducer such as a microphone, phonograph pickup, radio antenna, or 
strain gage to a level which is adequate for the operation of another 
transducer such as a loudspeaker or recording device. For example, the 
amplifier in a public address system must raise the signal level of the 
few picowatts available from the microphone to the several watts of 
power required by the loudspeaker. The amplifier usually consists of 
several stages. A stage is composed of an amplifying device such as a 
vacuum tube or transistor along with its associated circuit components. 
The stages are usually connected in cascade ; that is, the output of one 
stage is connected to the input of the following stage. 

The final amplifier in the chain is required to furnish the necessary 
power to the transducer in the output. Therefore, this final amplifier is 
called a power amplifier. This power amplifier must operate at high signal 
levels to accomplish its purpose. If the power amplifier is a vacuum tube, 
it may require appreciable input voltage but very little signal power for its 
operation. Consequently, the amplifiers which precede the vacuum tube 
power amplifier are usually called voltage amplifiers. On the other hand, 
if the power amplifier is a transistor, it may require appreciable current but 

240 



Small-Signal Amplifiers 241 

very little input power for its operation. The preceding amplifiers may 
then be referred to as current amplifiers. 

Both current and voltage amplifiers may operate at very low power 
levels, so the efficiency of the amplifier is not usually of prime concern. 
Also, the signal levels are small compared with those of the power amplifier. 
These factors make it possible to design amplifiers which have negligibly 
small distortion. Under these conditions the equivalent circuits may be 
used with a high-degree of accuracy. This type of amplifier will be known 
as a small-signal amplifier in this treatment. 

There is no intentional frequency selection in the untuned amplifier. In 
fact, an absence of frequency discrimination is desirable. This lack of 
discrimination is in contrast with the tuned amplifier which is usually 
designed to select a band of frequencies and reject those not lying within 
the desired band. Small-signal tuned amplifiers are discussed in Chapter 8. 

The equivalent circuits will be used to predict the response of vacuum 
tube and transistor amplifiers for various types of excitation. The most 
direct approach would be to draw a complete equivalent circuit of the 
amplifier and its coupling device; then derive an equation for the transfer 
function which is the ratio of the output quantity, or response, to the input 
quantity, or excitation. This transfer function would be a function of the 
complex frequency s. 1 As a result, the complete response could be obtained 
for any specified excitation or driving function. The response as a function 
of time could be obtained for any given excitation by obtaining the inverse 
transform of the product of the transfer function and the j-domain 
excitation. The steady-state frequency response could be obtained by 
replacing s by jm in the transfer function. 

One major problem in the procedure just outlined is the complexity of 
the equivalent circuit which results when the stray capacitance of the 
circuit is included. This problem can be circumvented by resolving it 
into two parts. First, the transfer function may be obtained for the "low- 
frequency" equivalent circuit in which the small stray capacitance and 
inductance may be neglected. Next, the transfer function may be obtained 
for the "high-frequency" equivalent circuit, in which the large series 
capacitors become short circuits and large shunt inductors become open 
circuits. Then the transfer functions of these two simplified equivalent 

1 A complete treatment of the complex frequency concept is given in many circuit 
texts among them, Network Analysis, Second Edition by M. E. Van Valkenburg, 
Prentice-Hall, Englewood Cliffs, New Jersey, 1964. The reader who is unfamiliar with 
this concept can understand the material which follows, in a restricted sense, by sub- 
stitutingyco for s in the equations and sinusoidal steady state in place of s domain in the 
written material. When nonsinusoidal waveforms are encountered, the response 
characteristics must be accepted without understandable proof unless the reader is 
familiar with the Laplace Transformation. 



242 



Electronic Engineering 



circuits may be combined to provide a complete transfer function of the 
amplifier stage. The transfer function of an amplifier which contains 
several stages is the product of the transfer functions of the individual 
stages, providing the amplifying devices (transistors, tubes, etc.) provide 
good isolation between the output and input of each stage. 



7.1 THE R-C COUPLED AMPLIFIER 

The resistance-capacitance coupled amplifier will be used to illustrate 
the foregoing procedure for determining the response of an amplifier. 
Typical R-C coupling circuits for both transistors and tubes are shown in 





(b) 

Fig. 7.1. Typical resistance-capacity coupled amplifiers, (a) Transistor amplifier; 
(6) tube amplifier. 

Fig. 7.1. In the R-C coupled amplifier the signal components of voltage 
and current are transferred to the following amplifier or load while the 
bias components are blocked by the capacitor C c . This R-C coupling, which 
is frequently known as resistance coupling, is a very commonly used 
method of coupling for amplifiers. 

In the discussion which follows, the assumption is made that the emitter 
(or cathode) is perfectly bypassed over the useful frequency range of the 
amplifier. Then the frequency characteristics of the amplifier will depend 
on the coupling capacitor C c and the stray circuit capacitance, but not on 
the emitter bypass capacitance. The validity of this assumption for a 
well-designed amplifier will be demonstrated in a following section. 

Equivalent circuits for the amplifiers of Fig. 7.1 are drawn in Fig. 7.2. 
In the transistor amplifier equivalent circuit the bias resistors R s and R 3 



Small-Signal Amplifiers 



243 




Fig. 7.2. Equivalent circuits for the amplifiers of Fig. 7.1. (a) Transistor; (b) tube. 

of Fig. 7.1 were combined into their equivalent parallel resistance R b . 
These equivalent circuits of Fig. 7.2 can be further simplified by combining 
the parallel resistances shown. Then, if the input circuits of the two 
equivalent circuits are omitted, the remaining simplified equivalent circuit 
can represent either the transistor amplifier or the tube amplifier. This 
simplified equivalent circuit is presented in Fig. 7.3. From Fig. 7.2, it may 
be seen that when the transistor circuit is being considered, the current 
source i\ = h /e i b , Y t is the sum of the load conductance 1/1?,. and the 
transistor output admittance (approximately h M ), and Y r is the sum of the 
bias conductance 1/.R,, and the input conductance (l/R,) of the following 
transistor (approximately l//i ie ). Observation also reveals that when the 




■.ccfuo*^ 



-cr- "tw 

/~ c 



Fig. 7.3. The single equivalent circuit which results from the simplification of either 
circuit of Fig. 7.2. 



244 Electronic Engineering 

vacuum tube is being considered, i x = g m v ak , Y t is the sum of the load 
conductance \jR p and the plate conductance g v = \\r v , and Y r is the grid 
circuit conductance 1/R,, of the following tube. 

The objective in this case is to obtain the output voltage v in terms of 
the forcing current i v In order to obtain a general solution without 
using differentials, the s-domain currents, voltages, admittances, and 
impedances will be used. Capital letters will be used to indicate these 
5-domain currents and voltages, and the complex frequency s will replace 
the frequency jco in the impedance and admittance terms. Writing s- 
domain 2 nodal equations for the circuit of Fig. 7.3, we find that 

W + sQV.-sCV.^-l! 
-sCV a + (Y r + sC)V o = 

Solving for V (using determinants), we find that 

v sCIj. 

° (y, + sC)(Y r + sC) - s 2 C 2 



(7.1) 



(7.2) 



but 

where 

then 



V " ~ ~ h Y lYr + (Y l+ Y r )sC (73) 

F --^T7, , , r,r. (7 ' 4) 

(5". + Y T )C 



YlYr 



= Ri + K (7.5) 



R t = ± and /? r = i 

K = ~t s + ll(R l + R T )C (? - 6) 

where Y t = y, + Y r . Or in alternative form 

F -- ,| "- . + W + JU c <7J) 

2 As noted in the List of Symbols (p. xx), lower case letters with lower case subscripts 
are used to represent time-varying components of voltages and currents. Thus, v is 
understood to imply v (t). Capital letters with lower case subscripts are used to represent 
5-domain voltages and currents. Thus, V implies V (s). (Since jco is a special case of 
s(s = a +ja>), V may also imply VjLjm) if the steady-state sinusoidal problem is being 
considered.) The reader should be especially careful not to intermix j-domain and 
time-domain terms. 



Small-Signal Amplifiers 245 

where 



R,h = 1/n 



let 



1 (7.8) 



(It, + R r )C 

tt) t is then the angular frequ ency at which the reactance of the coupling 
capacitor has the same magnitude as the total resistance in series with it 
(R, + R r ) because from Eq. 7.8 

R l + Rr = ^- (7.9) 

cojC 
Equation 7.7 may then be written 

V =-hR sh ~r- (7-10) 

In a vacuum tube amplifier where I x = g m V gk 

V =-g m V gk R ah -^— (7.H) 

The voltage transfer function, or voltage gain is 

G r = ^=-g m R„-^— (7.12) 

V ak S + Wi 

When the complex frequency s becomes large in comparison with co t , the 
transfer function (Eq. 7.12) reduces to 

G v ~-g m R sh =-K v (7.13) 

This K v of Eq. 7.13 is commonly known as the reference voltage gain of the 
amplifier. Sometimes K v is also known as the mid-frequency voltage gain. 
As previously mentioned, when the transistor amplifier is considered, 
the current 7 X of Eq. 7.10 is h u I b and Eq. 7.10 becomes 

V =-h fe hR sh -^— (7.14) 

S + «! 

In the transistor, the current transfer function or current gain is usually 
of greater interest than is the voltage gain. The current gain may be 
easily obtained after making the substitution 

V = I R t (7-15) 

where I is the current which flows through the input resistance R { of the 
following transistor. Substituting Eq. 7.15 into Eq. 7.14, we have 

V*i=-MA*-r— ( 716 > 

s + ft>i 



246 Electronic Engineering 

and 

g '-7--^ s tt- < 7 - 17 > 

l b K, S + C0 1 

As the frequency s becomes large in comparison with co u Eq. 7. 17 becomes 

G .~_>h£s* = _ K( (718) 

R i 

where A, is the reference current gain of the transistor amplifier. 

Equations 7.12, 7.13, 7.17, and 7.18 indicate that the transfer function 
of an R-C coupled amplifier (neglecting stray capacitance and inductance) 
may be written 

G--K— *- £rM (7.19) 

S + C0 1 K ~~*S 

where K is the reference (or mid-frequency) voltage or current gain of the 
amplifier under consideration. As noted before, the negative sign appears 
because of the polarity reversal experienced in the common-emitter 
amplifier. In other configurations, the negative sign would not appear. 

PROB. 7.1. A typical small-signal transistor has h u = 1 KO, h fe = 100, 
h re = 5 x 10~ 4 , and h oe = 5 x 10~ 5 mhos, at a given operating point. Several 
of these transistors are used in cascade with d-c load resistors R c = 5 KC1. 
Neglecting the loading effect of the bias resistors and the reactance of the 
coupling capacitors, verify the assumptions that R c^\jh oe and R t ~h,.. 
What are the percentage differences between R t and h ie and between the actual 
output admittance G and h oe l 

PROB. 7.2. Develop an expression for the reference voltage gain of a transistor 
amplifier. Answer: G v = h,RJ(h r h f R sh - 1). 

7.2 SAG AND LOW-FREQUENCY RESPONSE 

Sinc e the transfer functio n is defined as the ratio of the j-domain response 
to thefj-domain excitatiopi the ^-domain voltage response is 

(7.20) 

Then the s-domain response|may be obtaine d for any spec ific excitation. 
For example, let us consider ^negative step input (excitation ) of magnitude 
M. Since the Laplace transform of a step function of magnitude M is 
Mjs, the j-domain response of an R-C coupled amplifier to this step 
function is 




= [-K )=MK — — (7.21) 



s + coj 5 + co 1 



Small-Signal Amplifiers 

MK 



0.37 MK - 



247 




J- = (if, + R r )C 



Fig. 7.4. Time response of an R-C coupled amplifier to a negative step input of magni- 
tude M. 

Taking the inverse transform of each side of Eq. 7.21, we find that 

v = MKe- mii (7.22) 

A sketch of the output voltage v as a function of time is given in Fig. 

7.4. The output approaches zero as time approaches infinity. This 
response is to be expected because a step voltage is merely a d-c voltage 
that is switched into the amplifier at the time of reference, t = 0. Since 
the coupling capacitor blocks d-c, the steady-state output must be zero. If 
the step input voltage had been of positive polarity, the polarity of the 
output voltage would, of course, be reversed. 

Although the R-C coupled amplifier is not suitable for amplifying d-c 
signals, the information gained from the step function analysis is very 
useful in predicting the behavior of the amplifier when it is called on to 
amplify rectangular pulses. A series of rectangular pulses is shown in Fig. 

7.5. These pulses have amplitude M, pulse duration d, and period (of 
repetition) T. In the case d = 7/2, this signal is known as a square wave. 

If we assume the excitation to be a series of voltage pulses as illustrated 
in Fig. 7.5, we may express the time-domain input voltage as the sum of 
a series of step voltages. 

», = M[u(t) - u(t -d) + u(t - T) - u(t -(T+ d)) + u{t - IT) ■ • •] 

(7.23) 



M 



< T > 

d <- 



T T+d 

t » 



2T 2T+d 3T 3T+d 



Fig. 7.5. A series of rectangular pulses. 



248 Electronic Engineering 

The u(t — a) symbolizes a step voltage that occurs at t = a and so on. 
The Laplace transform of this excitation voltage is 



V t = m(- - — + — - + - ) (7.24) 



l1 s 

s s 



The i-domain output voltage is 

/ 1 p-ds p-Ts p -(T+d)s -2Ts \ 

V = -MK\—± - + - e - + 

\S + C0 1 S + fDj 5 + C0 1 S + (X> x s + co 1 I 

(7.25) 
The time-domain response is obtained from the inverse transform 3 

v = -MK[e~ mit - u(t - d)e-'° lU - d) + u(t - T)e~'° lU - T) 

-u(t-T- d)e"° lU - T ~ d) + u{t - 2T)e- m ^- 2T) . . .] (7.26) 

The time response may be plotted by adding the exponential terms of 
Eq. 7.26, as illustrated in Fig. 7.6. Since the average steady-state current 
through the coupling capacitor must be zero, the cross-hatched area labeled 
A below the reference axis must approach the corresponding area labeled 
B above the axis after several periods T have elapsed. 

Ideally, the output voltage should remain constant during the pulse. 
However, the output current and voltage decays exponentially as shown 
because of the charging of the coupling capacitor. The departure of the 
actual o utput from the id eal flat response at the term ination of the pulse 
i s called sag, hrom a consideration of the first pulse as shown in Fig. 7.6, 
the sag can be determined by solving Eq. 7.26 at t = d. Then 

Sa gl = KM{\ - e-° ld ) (7.27) 

The fractional sag, or ratio of sag to the output at the beginning of the 
pulse, is of greater significance than the amount of sag. 

Fractional sag = -5ii = 1 - e ~ aid (7.28) 

KM 

If the pulse duration d is small in comparison with l/coj = (R [ + R r )C, 
the fractional sag can be determined more easily than by the use of Eq. 
7.28. The initial slope of the exponential decay is, using Eq. 7.26, 



Initial slope = — 

dt 



= KM Ml (7.29) 



(=0 



The exponential sag closely follows the initial slope as long as the pulse 

"This time transformation is treated in Network Analysis, Second Edition, by M. E. 
Van Valkenburg, Prentice-Hall, Englewood Cliffs, N.J., 1964. 



Small-Signal Amplifiers 

Function 

-KMe-"!' 



249 



Plot 



KMu(t-d)e- a i('- d > 



-KMu(t-T)e- w i<'- T > 



KM u(t-T- d)e-<*i (t-T-d) 




T+d t >■ 



Sum of the 
above functions 



-KM^r-S. 









^=5?^-^— 






Sag 










<* 



«T" 



d i T T+d 2T 

Fig. 7.6. The time response of an R-C coupled amplifier to a periodic rectangular pulse. 

duration is small (0.1 or less) in comparison with the time constant l/co^ 
Then the total sag is approximately equal to the pulse duration times the 
initial slope 

Sag ~ KMa) r d (7.30) 

and 

Fractional sag ~ <a x d (7.31) 

Equation 7.31 points up the desirability of a small value of <o l when 
rectangular pulses are to be amplified. 

Figure 7.6 shows that the first pulse after / = has greater sag than the 
succeeding pulses. The sag decreases somewhat as the transient dies out. 
However, the fractional sag remains constant if the pulse amplitude is 
measured from the base line. 



250 



Electronic Engineering 



The excitation and response quantities were assumed to be voltage 
pulses in the preceding discussion. Current pulses could have been con- 
sidered with equal validity since the transfer function was developed 
generally for either case. 

PROB. 7.3. A periodic rectangular pulse of 1 msec duration is applied to the 
input of an R-C coupled amplifier. What must be the value of (o 1 to provide a 
fractional sag of 0.05 or 5 % ? Answer: co 1 ~ 50. 

PROB. 7.4. The R-C amplifier of Prob. 7.3 has R, = 10 Kfl and R r = 2 KO. 
What coupling capacitance is needed ? 

An expression of the gain as a function of frequency may be obtained by 
replacing s by jco in Eq. 7.19. A plot of G as a function of co is known as 



0.707K 




O.lW! 



Fig. 7.7. Amplitude and phase response at low frequencies. 

the frequency response curve of the amplifier. This frequency response is 
more accurately described as the steady-state response to sinusoidal 
excitation. 



'G(jco) = -K 



G(jco) = -K 



](Q 



jco + w 1 

1 



= -K 



(7.32) 



(7.33) 



I-Kcojco) l-j(fjf) 

where <Wj = 2-rrf^ Figure 7.7 is a plot of both phase and magnitude of 
G(jco) as a function of to. It is evident that Eq. 7.33 will yield the 
reference gain when co is large compared with a^. The upper end of the 
low-frequency range is defined as 10 eo x . Above this frequency the gain 
does not differ from K by more than 1 % and the phase shift through the 
amplifier is approximately 180°. The frequency and phase response- plots 
are made on semilog-ruled paper. 

PROB. 7.5. Write the gain (Eq. 7.33) in polar form for values of to = 0.1 to^ 
(t> = coj, and co — 10 coj. Compare the results with the plots in Fig. 7.7. 
Answer: At to = 0.1 m ly G = {Kjl0) \-™° . 



Small-Signal Amplifiers 



251 



PROB. 7.6. Determine the value of <o 1 for a vacuum tube amplifier which has 
fi = 20, /■„ = 10 KQ, R P = 50 KQ, *,, = 0.5 megohm, and C c = 0.2 /if. What 
would be the value of /i ? 

In a pole-zero plot of Eq. 7.19 given in Fig. 7.8, the scale factor is — K. 
The frequency and phase response can be obtained directly from the pole- 
zero plot. For example, the distance from the origin to the specific value 
of co is the magnitude of the numerator of Eq. 7.32. In addition, the angle 
between the positive real axis and theyw axis is the angle of the numerator. 




Fig. 7.8. A pole-zero plot for the R-C coupled amplifier. Stray capacitance has been 
neglected. 

The distance between the pole and the specific en is the magnitude of the 
denominator of Eq. 7.32. The angle of the denominator is the angle <f> 
between the real axis and the line drawn between the pole at e^ and to. 
Therefore, the phase angle of the fraction is the difference between these 
two angles which is d in the figure. Note that when s = ja) lt 6 = 45°. 
The pole on the negative real axis indicates an exponentially decaying term 
in the transient response. 



7.3 THE MID-FREQUENCY RANGE 

The term " mid-freque ncies" is used to designate the band of f requencies 
for which all the reactanc es in the circuit are negligible. In the mid- 
frequencies the reactances of the coupling capacitors and bypass capacitors 
are so small that these reactances are assumed to be zero. Consequently, 
the lower limit of the mid-frequency range is defined as the frequency 
which is ten times co x . Also, at mid-frequencies the reactance of the stray 
shunt capacitance, which is not shown in the circuit diagram, is so large 
that this reactance is assumed to be infinite. 



252 



Electronic Engineering 



The graphical analyses presented in the preceding chapters were based 
on the assumption that the resistor which is connected between the collector 
(or plate) and the power supply is the total load resistance on the amplifier. 
This resistor has been designated as R L . However, it now becomes 
apparent that the load impedance of an R-C coupled amplifier decreases 
as the frequency increases toward the mid-frequency range. This a-c 
load impedance Z L is represented in the equivalent circuit of Fig. 7.9 by 
the elements inside the rectangular enclosure. In the mid-frequency range 
the load impedance becomes essentially a pure resistance r L . This a-c 
load resistance r L is the parallel combination of the d-c resistance R L and 
the resistance R r , as may be seen from Fig. 7.9. 




Fig. 7.9. The a-c load impedance of an R-C coupled amplifier. 

The quiescent operating point of an amplifier must lie on the d-c load 
line which is determined by the d-c load resistance R L and drawn in the 
manner described in the preceding chapters. However, in the mid- 
frequency range the time-varying current-voltage relationships of the 
collector circuit are determined by the a-c load resistance r L . Therefore, 
an a-c load line can and should be drawn on the collector (or plate) 
characteristics to graphically display the dynamic operation of the ampli- 
fier. This a-c load line has the slope —\\r L and must pass through the 
quiescent operating point. The a-c load line may be readily drawn if a 
point in addition to the Q point is located. The x axis intercept may be 
located, as shown in Fig. 7.10, from the relationship 

Av L = M L r L (7.34) 

When the Q-point value (7 C or I P ) is chosen for &i L , then Av L will be the 
voltage difference between the g-point voltage and the a>axis intercept of 
the a-c load line. 

Since the a-c load impedance of an R-C coupled amplifier is less than the 
d-c load resistance, the a-c load line has a steeper slope than the d-c load 
line. In the transistor amplifier, the a-c load line is usually much steeper 
than the d-c load line because of the relatively low input resistance of the 



Small-Signal Amplifiers 



253 



following transistor. This relatively steep slope of the transistor a-c load 
line is desirable because the objective in the transistor amplifier is to have 
maximum signal current transferred through the coupling capacitor to the 
following transistor or load. On the other hand, the vacuum tube is a 
voltage-operated device. Therefore, the ideal a-c load line would have the 
same slope as the d-c load line. Then the maximum voltage gain and 
maximum voltage swing would be achieved, as may be seen from Fig. 
7.10Z>. However, the resistance between the control grid and the cathode 
would have to be infinite in order to make the a-c load line coincide with 



a-c load line 




d-c load line 




A£i Aw, 



(a) (6) 

Fig. 7.10. Typical d-c and a-c load lines for R-C coupled (a) transistor and (b) triode 
tube circuits. 

the d-c load line. Therefore, the grid resistor R g is usually chosen in 
accordance with the maximum value recommended by the tube manu- 
facturer. 

The maximum permissible value of grid circuit resistance R g depends on 
the characteristics of the tube and the completeness of the evacuation 
process. Since it is practically impossible to completely evacuate an 
enclosure, many molecules remain in a satisfactorily evacuated tube. Some 
ionization of these molecules occurs during normal operation of the tube. 
The positive ions are accelerated toward the grid because it is the most 
negative electrode. The grid may then become positive with respect to 
ground because the positive ions remove electrons from the grid and cause 
a small current flow in the grid circuit. If the grid circuit resistance is too 
high, the bias may be appreciably reduced by the IR drop in the grid 
circuit. This reduction of grid bias may cause serious consequences in a 
power amplifier which is normally biased at a point which fully utilizes 
the power dissipation capabilities of the tube. A significant reduction in 
bias from the design value then causes increased plate temperature and 
expulsion of gas from the electrodes. This additional gas causes increased 



254 Electronic Engineering 

ionization which results in increased grid current and a further reduction 
of bias. This cumulative chain of events will soon destroy the tube. 

Electrons which leave the cathode may strike the grid even when the 
grid is a volt or so negative with respect to the cathode. The current 
which results is opposite in direction to that caused by ionization in the 
tube. In certain types of tubes such as high fi triodes, this contact- 
potential current is large compared with that due to ionization; first, 
because the operating bias is of the order of a volt and second, because the 
ionization in the tube is small compared with that of a power amplifier. 
This grid current reversal results from the smaller grid bias voltage and 
lower operating temperatures. 

The electrons collected by the grid will not affect the bias appreciably 
if the grid circuit resistance is of the order of one megohm or less. How- 
ever, if the grid circuit resistance is several megohms, a bias voltage of the 
order of one volt may be developed. This method is sometimes used to 
bias a high /j, triode. 

In the R-C coupled amplifier, the quiescent operating point is not 
usually located at the center of the d-c load line. The optimum location 
of the Q point is at least partially dependent on the value of the a-c load 
resistance. The design of an amplifier usually begins with the final load 
and then proceeds toward the input of the amplifier. The quiescent 
operating point of the amplifier is usually determined by the current and 
voltage requirements of the a-c load. The d-c load line can then be drawn 
through the Q point to the collector supply voltage. The a-c load can then 
be computed and the a-c load line drawn. This graphical construction 
provides visualization of the operation and linearity of the amplifier. 

PROB. 7.7. A 2N1 92 transistor must furnish 4 ma peak-to-peak exciting current 
to a following transistor which has h ie = 200 a and shunting bias resistance 
R b = 500 £2. Choose a suitable Q point and calculate the d-c load resistance for 
Vcc = —15 v. Draw both the d-c and a-c load lines. Calculate the reference 
current, voltage, and power gains of the amplifier. 

PROB. 7.8. The amplifier of Prob. 7.7 has V cc raised to -25 v. Using the same 
Q point as Prob. 7.7, determine the d-c load resistance. Compare the current 
gain and temperature stability of the Q point for the two values of V cc . 

7.4 SHUNT CAPACITANCE IN AMPLIFIERS 

Capacitance exists between the various parts of any circuit. Also, the 
conductors used to connect the parts have inductance. At low and middle 
frequencies the effect of this "stray" capacitance and inductance is 
negligible, but at high frequencies their effect must be taken into account. 
The high frequency range of an R-C coupled amplifier is the range in 



Small-Signal Amplifiers 



which the shunt capacitance affects the gain 
which may limit the high frequency gain are 
The reactance of the connecting wires is 
useful range of an R-C coupled amplifier. 

The vacuum tube has capacitance between 
these electrodes consist of conductors 
insulating material. The magnitude of 
listed in most tube manuals. These interelectrode 
order of a few picofarads. Transistors 
base and each of the other two regions 



USUI 



its various electrodes because 

bparated by vacuum or other 

ch interelectrode capacitance is 

capacitances are of the 

have capacitance between the 
This is the junction capacitance 



eac 



also 



vn 



13 f 



Cte 



C11 



=r CD =r 



Cl3' 



B,h 



Vol 



Fig. 7.11. Shunt capacitance in an 



In 



discussed in Chapter 3. As noted, junction 
not only on the cross-sectional area of the junction 
voltages of the transistor. Additional 
transistor amplifiers will be discussed late} - 

Figure 7.11 is a high-frequency equivalent 
coupled amplifier. The diagram includes 
represent the distributed circuit capacitances 
capacitances of the amplifying device, 
resistance of each amplifier is shown a« R 
are not shown because their effects are usuai 
for which the shunt capacitances are effective 

Figure 7. 1 1 will be used in determining 
at various points in the circuit. The input 
amplifier 1 will first be considered. This 
the input current i x can be expressed as a 
Referring to Fig. 7.11, and using j-domaih 



A = h + h = sC xx V iX + 

where C u is the effective capacitance betw 
is the effective capacitance between the 



255 

of the amplifier. Other factors 

discussed in a following section. 

ally negligible throughout the 




R-C coupled amplifier. 



capacitance is dependent 
but also on the electrode 
high-frequency limitations of 
in this chapter. 

circuit for a two-stage R-C 

lumped capacitances which 

as well as the interelectrode 

addition, the effective shunt 

sh . The d-c blocking capacitors 

lly negligible at the frequencies 



the effective shunt capacitance 

or driving point admittance of 

admittance may be determined if 

function of the input voltage v iV 

variables, we find that 



sC 12 (V (1 V gl ) 



(7.35) 



een the input terminals and C ia 
input and output terminals of 



256 Electronic Engineering 

amplifier 1. But 

Ki = G vl V a (7.36) 

where G vl is the voltage gain of amplifier 1 . Then 

h = s[C n + C 12 (l - G vl )]V n (7.37) 

Y t = £- = s[C u + C 12 (l - G vl )] (7.38) 

' ti 

The effective input capacitance of the circuit appears to be the bracketed 
term in Eq. 7.38. Notice that the mutual capacitance C 12 has made a 
contribution which is (1 — G x ) times as great as if it had been connected 
directly across the input terminals. In a common emitter amplifier, Gj 
contains a negative sign. Then, for a common emitter amplifier the effective 
input capacitance would be 

Qn = C X1 + C 12 (l + UteGil) (7.39) 

where ReG 1 refers to the real part of G vl . In case C u and C 12 were of the 
same order of magnitude, the effect of C 12 would completely dwarf C n *n 
determining the input capacitance of a high-gain amplifier. 

In the usual case, when the load impedance is not purely resistive, the 
effect of C 12 is not purely capacitive but includes either a positive or 
negative conductance term depending on the phase angle of the load. This 
fact may be verified in the following problems. 

PROB. 7.9. If the input signal is V sin (5 x 10 5 )f, calculate the input admittance 
of a common cathode triode amplifier which has C gk = 5 pf, C tp = 4 pf, 
and G v = 50H35!. Determine the effective input capacitance. Answer: 
Yi = (1.03 x ;o- 4 )i«< 

PROB. 7.10. What is the conductance and susceptance of the amplifier of Prob. 
7.9 when G„ =50la?s!? 

It may be seen from Fig. 7.11 that the total effective capacitance in 
parallel with the effective shunt load resistance, R sh , is the sum of the 
effective output capacitance of amplifier 1 ; the distributed wiring capaci- 
tance, C w , and the effective input capacitance of amplifier 2. Again, the 
effective output capacitance of amplifier 1 may be determined by finding 
the sum of the currents which would flow through C 12 and C 13 as related 
to the output voltage v ol . This sum, in the s domain, is 

/.i = sC 13 V 01 + sC 12 (V 0l - V a ) (7.40) 

but, for an active amplifier, V n = V o1 /G v1 . Then 

i„-.[c+c„(i-£); 



V„ (7.41) 



Small-Signal Amplifiers 257 

Since G vl is normally large in comparison with unity, the term ljG vl may 
be neglected. Consequently, the effective output capacitance is approxi- 
mately C 13 + C 12 . The total shunt capacitance in parallel with R sh is then 

C, h = C 13 + C 12 + C w + C j2 (7.42) 

where C <2 is the effective input capacitance of amplifier 2. From Fig. 7. 1 1 
and Eq. 7.38 

C« = C 21 + C 22 (l - \ReGJ) (7.43) 

where ReG v2 is the real part of the transfer function G vi (jco). 
As a result, the total effective capacitance in parallel with the load resist- 
ance of amplifier 1 is, within the useful range of the amplifier, approxi- 
mately 

C sh = C 13 + C 12 + C. + C 21 + C 22 (l - \ReGJ) (7.44) 

The reactance of capacitance C sh is part of the load impedance of ampli- 
fier 1 . The effect this reactance has on the response of the amplifier will be 
discussed in detail in the next section. However, some casual observations 
should be made at this point. First, the gain of an R-C coupled amplifier 
will decrease as the frequency increases because of the reduction in load 
impedance caused by the effective capacitance in parallel with the load 
resistance. Second, the effective input impedance of an amplifier becomes 
part of the load impedance of the preceding amplifier and therefore affects 
only the transfer function of the preceding amplifier. Similarly, the input 
impedance of the first amplifier stage becomes part of the load of the 
exciting source for that stage. Consequently, the input impedance of any 
stage does not affect the transfer function of that stage ; but it does affect 
the character of its excitation. 

At this point, the observing reader may be seized with a mild case of 
panic as he becomes aware that the transfer function of any amplifier is 
dependent on the input impedance and consequently the transfer function 
of the following amplifier. This condition, known as "interacting" 
stages, requires that a simultaneous solution be made of the entire ampli- 
fier rather than considering it one stage at a time. In addition, the negative 
conductance (or resistance) term which appears in the input admittance 
(or impedance) when the amplifier has an inductive load should cause 
concern over the stability of the amplifier because sustained oscillations 
frequently occur in devices which exhibit a negative resistance character- 
istic. 

From the foregoing discussion and problems, it should be evident that 
capacitance between the input and output circuits of an amplifier may be 
very detrimental at high frequencies. As indicated in Chapter 6, the 
pentode tube is frequently an easy solution to the problem when the 



258 



Electronic Engineering 



amplifier is called on to handle high frequencies. Also, transistors which 
have extremely small collector-base capacitance are available. The 
grounded grid or common base amplifier configurations may also be used 
to advantage under certain circumstances because the grid or base region 
tends to shield the input (emitter) circuit from the output (collector or 
plate) circuit. A technique known as neutralization is also used in some 
cases to eliminate the effects of the input-output capacitance. Essentially, 
neutralization is accomplished by providing a voltage 180° out of phase 
with the plate or collector voltage. An appropriate capacitance from this 
voltage point to the input circuit will cancel the effects of the initial 
input-output capacitance. This technique is discussed in detail in Chapter 
8 in conjunction with tuned amplifiers. 



7.5 AMPLIFICATION AT HIGH FREQUENCIES (SINGULARITIES 
FAR REMOVED FROM THE ORIGIN) 

An equivalent circuit which is suitable for both high and mid-frequency 
amplification is given in Fig. 7.12. This circuit is based on the low- 
frequency equivalent circuit of Fig. 7.3. In the high-frequency circuit the 



»2-*- 



»2-*- 



<P T* i*' T c " i* Tc/° <P 



5 \ 


>Rsh 


O 

"c sh 

o 



(a) 



(b) 



Fig. 7.12. High-frequency equivalent circuit for transistor or vacuum tube R-C coupled 
amplifier. 

coupling capacitor has been removed because its reactance is negligible. 
In addition, the equivalent shunt capacitors have been added. The circuit 
of Fig. 7.126 has been further simplified by the combining of the parallel 
resistances into one equivalent resistance R sh and the parallel capacitance 
into one equivalent capacitance C sh . With this simplification, the output 
voltage is 

-h _ ~h 

Y sC sh + l/R sh 
Finally 

-h 1 



V n = 



(7.45) 



V n = 



C sh s + (l/R sh C sh ) 



(7.46) 



Small-Signal Amplifiers 259 

Now, let ft> 2 be the frequency at which X CUM = R sh . 
Then 

a> t = — — (7-47) 

and 

V = =^— — (7.48) 

C sh s + w 2 
or 

V ° = ^ -T '■*- "f^ (749) 

R*h C sh s + w 2 s + w 2 

For a vacuum tube amplifier J 2 = g m V g and 

G. = £ = -* m K rt "f 8 - (7-50) 

K 9 s + w 2 

or 

0.= -*-^ (7.51) 

s + <w 2 

Similarly, for a transistor amplifier / 2 = /i /e /i,i and 7 b2 = VJR i . Then, 

r = — = — fe ^ sh M2 = —K ( ° 2 (7 52) 

'hi R i s + o>2 s + m 2 

The time response may be obtained for any exciting function which may 
be specified in the s-domain. For example, to obtain the time response to 
a step voltage input, let V g = — Vjs. Then 

V = ^ = VK ° m "- = d + g ( 7. 5 3) 

s s(s + a> 2 ) s s + ft>2 

where ^ = VK V and 5 = - Ftf„. Then 

v = VK,(1 - «-") (7.54) 

A sketch of the output as a function of time is presented in Fig. 7.13. 
The output voltage rises exponentially toward the value VK V . The "rise 
time" is, by definition, the time required for the voltage to rise from 10% 
of VK V to 90% of VK V . This rise time is about 2.2/a> 2 or 2.2R ah C sh . 

PROB. 7.11. An amplifier has R, h = 50 Kfi and C, h = 100 pf. What will be 
the rise time of the response if a unit step function is applied to the input? 
Answer: 11 usee. 

An expression for the steady-state frequency response may be obtained 
by letting s = jo>. Then Eq. 7.52 becomes 

G(,V) = -K— 5— = -K I (7.55) 

JO) + 0) 2 1 + J(0)l0) 2 ) 



260 



Electronic Engineering 



' ' 0.63 VK„ -— 




Time- 



Fig. 7.13. The effect of shunt capacitance on the time response. 




0.1«2 <»>2 10<<>2 
o> *- 



+90 



0.1«2 ( ^l 10(02 



Fig. 7.14. Frequency and phase response at high frequencies. 




Fig. 7.15. Pole-zero plot for high-frequency case of an R-C coupled amplifier. 



Small-Signal Amplifiers 



261 



A sketch of gain and phase as a function of frequency is presented in Fig. 
7.14. The high-frequency range is denned as these frequencies above 0. 1 co 2 . 
A pole-zero plot of Eq. 7.52 (or 7.51) is given in Fig. 7.15. In this plot, 
there is a pole at — eo 2 , but there are no zeros except at infinity. The 
numerator in Eq. 7.52 is a constant (<w 2 ) ; therefore, the amplitude response 
at real frequencies is proportional to u> 2 divided by the distance from a 
specific value of s on the jto axis to the pole at —<a 2 . In this case, the 
scale factor has a value of K. 



-*- 9- 



—0)2 



-Wl 



(a) 



0.7 K 




^Amplitude 



^ Phase 



0.1 oil 



wi 



10o>i 




(b) 



Fig. 7.16. Characteristics of an R-C coupled amplifier, (a) A pole-zero plot of the R-C 
coupled amplifier; (b) frequency response and phase shift of an R-C coupled amplifier. 

The pole-zero plot for the high-frequency case (Fig. 7.15) and the pole- 
zero plot for the low-frequency case (Fig. 7.8) both represent the same 
R-C coupled amplifier. Therefore, a reasonable conclusion might be 
drawn that the poles and zeros of these two cases might be included in a 
single pole-zero plot, as shown in Fig. 7. 1 6a. This is a pole-zero plot which 
represents the entire R-C coupled amplifier, including the stray shunt 
capacitance. From this pole-zero plot and inspection of the transfer 
functions for the low-frequency and high-frequency cases, the following 



262 Electronic Engineering 

transfer function may be written for the R-C coupled amplifier. 

G=-K ^s (7. 56) 

(s + «i)(s + w 2 ) 

Notice that Eq. 7.56 reduces to the low-frequency equation 7.19 when s is 
small in comparison with co 2 . When s is large in comparison with co 1( 
Eq. 7.56 reduces to the high-frequency equation 7.51. Also when s is 
large in comparison with w 1 but small in comparison with a> 2 , G appro- 
priately reduces to —K. 

The complete frequency response of the R-C coupled amplifier may 
either be obtained by piecing together the results from the low-frequency 
and high-frequency equivalent circuits or by replacing s by jco in Eq. 
7.56. To show the detail of the variation of gain with frequency at both 
ends of the frequency spectrum, it is necessary to use a logarithmic 
frequency scale. However, since the ear responds logarithmically to both 
frequency and amplitude variations, a logarithmic frequency scale is 
realistic. The gain scale may be either logarithmic or linear. A complete 
frequency response curve as well as the complete phase response is drawn 
in Fig. 7.166. 

PROB. 7.12. An R-C coupled amplifier has R P = 50 KQ, R g = 1 megohm, 
r B = 10 Kfl, C c = 0.05 nf, C 01 = 10 pf, C i2 = 100 pf, C w = 20 pf and g m = 
2000 /<mhos. Sketch a frequency response curve for the amplifier. Indicate/i,/ 2 , 
K, low-frequency range, mid-frequency range, and high-frequency range. 
Answer: f x = 3. 16 Hz, K v = -16.5,f 2 = 148 KHz. 

The complete time response for any given input signal can be obtained, 
as previously discussed, by taking the inverse transform of the j-domain 
excitation and transfer function product. For example, the complete time 
response of the R-C coupled amplifier may be obtained with the aid of the 
complete transfer function Eq. 7.56 as follows. Let the input be a step- 
voltage having magnitude —M. 

Ve = MK sjo, = MK oh (? 

S (S + WjXs + tt) 2 ) (5 + WjXs + W 2 ) 

Using partial fraction expansion, we find that 

V = MK M ; -^_+-L- (7.58) 

(S + <0,XJ + W 2 ) S + ft)! S + W 2 

From an evaluation of A and B, 

_ MK(o 2 r — _ MKa h 



CO, — ft), 



Small-Signal Amplifiers 



263 



Take the inverse transform of Eq. 7.58, noting that co 2 is usually very large 
in comparison with ei^; then 



v ca MK(e~ 



-ait 



--M2* 



) 



(7.59) 



The plot of Eq. 7.59 is difficult to make for a well-designed amplifier 
because a time scale which is appropriate for showing the rise time will 
not illustrate the exponential decay caused by the coupling capacitor, 
and vice-versa. This problem is solved when the response is displayed on 
an oscilloscope by using a fast sweep to view the front edge response (rise 
time) and a slow sweep to view the exponential decay or sag. Figure 7.17 
is a sketch of the response v B as a function of time. In this sketch, the 
chosen ratio of a> 2 to a)j is small in comparison with a well-designed 
amplifier. 




Fig. 7.17. The complete time response of an R-C coupled amplifier to a step function 
input. 

As previously mentioned, the R-C coupled amplifier is commonly used 
to amplify periodic rectangular pulses. The time-response equation could 
be derived for a pulse input by the technique used to obtain Eq. 7.26. 
The derivation would be based on the complete transfer function Eq. 
7.56. However, in making a plot of the time response for these pulses, it 
would be almost impossible to display the rise times. Hence the complete 
response would appear almost identical to Fig. 7.6, assuming the excitation 
to be the same in the two examples. 

PROB. 7.13. Derive an expression for the time response of an R-C coupled 
amplifier to a periodic rectangular pulse of amplitude M, duration d, and 
period T. 

7.6 DIFFUSION CAPACITANCE 

The preceding discussion may have given the impression that the shunt 
capacitance of an amplifier is the only cause of deteriorated performance 
at high frequencies. This impression is approximately true for an R-C 



264 



Electronic Engineering 



coupled vacuum tube amplifier, but for a transistor, the junction capaci- 
tance is not strictly a shunt capacitance, as illustrated in the equivalent 
circuit of Fig. 7.18. In this figure the effective base resistance, r b , prevents 
either C x or C„ from being a pure shunt capacitance. Actually, the base 
resistance is distributed throughout the thin wafer of base material, and a 
different resistance would exist between each point on either junction and 
the base electrode. Consequently, the equivalent circuit of Fig. 7.18 is a 
gross approximation, but it is a great improvement over the assumption 
that C x and C 2 are in shunt with the emitter-base and collector-base 
terminals, respectively. As far as the collector junction capacitance C c is 
concerned, the resistance r„ produces the effect of a capacitor having loss. 



'1 r e 

— MAMA- 

Ci 



»2 



-vwvw- 



-It- 



r mle 

o 



>n 



Fig. 7.18. An approximate high-frequency circuit for a transistor. 



This resistive component will have little effect on the performance of an 
R-C coupled amplifier but will need to be included in the analysis of other 
types of amplifiers such as tuned amplifiers which will be considered in 
Chapter 8. 

In reference to Fig. 7.18, the capacitance C x is actually composed of two 
different capacitances: one is the emitter-base junction capacitance C e 
which has been discussed previously, and the other results from the carriers 
which are injected from the emitter into the base. The charge from these 
carriers is neutralized by oppositely charged carriers entering the base from 
the external circuit, as discussed in Section 4.4. For example, in a p-n-p 
transistor, the positive charge in the base which results from holes injected 
from the emitter is neutralized by negative carriers entering from the base 
electrode. This means that a positive charge will be transferred across the 
emitter junction whenever the emitter base voltage is increased, and this 
transfer will result in current flow in the base circuit to neutralize the charge. 
Conversely, a reduction in emitter-base voltage reduces the positive carriers 
in the base region. The extra negative carriers then flow back out the base 
lead. This transfer of charge into and out of the base region is analogous to 



Small-Signal Amplifiers 



265 



the charge and discharge of a capacitor. This phenomenon is accounted for 
by a capacitance known as diffusion capacitance, C D . 

With the aid of Fig. 7.19, we may develop an expression for the diffusion 
capacitance. The figure shows the distribution of positive charge in the 
base region. The carrier density resulting from injected carriers is p at 
the emitter junction (x = 0) when the emitter current is at the bias level. 
The carrier density at the collector junction depletion region (x = w) is 
essentially zero, as discussed in Section 4.4, and the carriers lost by recom- 
bination are neglected. Thus the charge density decreases linearly with 



Po + Ap 






c £ P 


*^H|V. „. c 




o J2 


^*2|||&V CD « 


c 
.2 c 


*- 0) 




•ffi.2 


1 ff 




S ™ 


E S 




a 


fi o 







Fig. 7.19. Charge distribution in the base of a p-n-p transistor. 



distance because dp\dx must be constant if the diffusion current is the same 
for all values of x. The density and distribution of negative carriers is 
approximately the same as that of the positive carriers, as discussed in 
Section 4.4. When the emitter junction voltage v E is increased by Ad £ , 
the carrier density in the base at the emitter junction is increased by Ap. 
Then, the increase in positive charge in the base is 



AQ = 



q Apw/4 



(7.60) 



where A is the cross-sectional area of the active base region. The diffusion 
capacitance is 



C n = 



M 

Av 



E 



But, in reference to Fig. 7.18, 

Av E = Aiy, 

And, using the diffusion equation (Eq. 4.16), we find that 

Ai x =qD p — A 

w 



(7.61) 
(7.62) 

(7.63) 



266 Electronic Engineering 

Then, substituting Eqs. 7.60, 7.62, and 7.63 into Eq. 7.61 we have 

Cn = r£- (7-64) 

It may be seen from Fig. 7.18 that the signal current to the transistor is 

h = h + h ( 7 - 65 ) 

If, for example, a small input voltage is applied to the transistor 

/. = ^ + sdV. (7.66) 

r e 

where V t is the voltage across the emitter-base junction as indicated in 
Fig. 7.18. If the frequency is low so that the capacitor current I 2 is 
negligible in comparison with I lt the current gain with the output shorted is 

/ I r 

«o = 7 = TT ( 7 - 67 > 

where a is identical with a as defined for the low-frequency circuits of 
Chapter 4. But in the more general case, the current gain with output 
shorted is 









a = 




hr e 




V e + sC x V e r e 


using 


the value of a in 


Eq. 


7.67, 


we have 




1 






a = 


«o 




1 + sC 1 r e 


Let w 


Then 
















a = 


<*o 




1 + s/co a 


or 








a = 


(On 
a 



s + (o a 



(7.68) 



(7.69) 



(7.70) 



(7.71) 



At the real frequency s =ju> x , |a| = 0.707a . This frequency m^, or 
preferably f x = (oJ2tt, is known as the "alpha cutoff" frequency. Then 



/- = r^— (7-72) 



where C 1 is equal to (C D + C e ). 



Small-Signal Amplifiers 267 

PROB. 7.14. A certain p-n-p germanium transistor has an effective base width 
w = 2 x 10 -3 cm. Calculate C D at I E = 0.5 ma. If C e is 10 pf, find the value 
of/ a . Answer: C D = 870pf,f x = 3.28 MHz. 

Transistors which have uniform doping in the base usually have values 
of diffusion capacitance which are large in comparison with the junction 
capacitances C e and C c . However, manufacturing techniques which 
minimize this diffusion capacitance have made possible the production 
of high-frequency transistors with values of diffusion capacitance of 
the same order of magnitude as the junction capacitances. One 
common type of high-frequency transistor is the graded-base or drift- 
field transistor. In this type, the net base doping is almost zero at the 
collector junction but the doping increases rather rapidly with distance 
toward the emitter junction. This graded doping produces a nonuniform 
distribution of mobile carriers which tend to diffuse toward the collector 
junction and thus establish an electric field in the base which inhibits the 
diffusion of the majority carriers but assists the diffusion of carriers which 
have been injected from the emitter. This "built in" electric field adds to 
the field which results from the nonuniform distribution of neutralizing 
carriers which are present whenever carriers are injected into the base from 
the emitter. 

The increased velocity of carriers in the graded base decreases the 
diffusion capacitance for a given collector current. Also, the collector- 
base junction capacitance is low because of the low doping concentration 
and wide depletion region at this junction. This wide depletion region 
also provides relatively high collector-base breakdown voltage. The 
relatively high doping concentration at the emitter junction requires that 
the emitter doping concentration must be unusually high for good emitter 
efficiency. These high doping concentrations result in low emitter-base 
breakdown voltage. 

7.7 THE HYBRID-* EQUIVALENT CIRCUIT FOR A TRANSISTOR 

All the h parameters become functions of frequency when the frequency 
is so high that the junction capacitances and diffusion capacitance can no 
longer be neglected. Also, these capacitances cannot be represented by 
simple lumped elements in the A-parameter circuit. Therefore, an equiv- 
alent circuit known as a hybrid-n has been developed and widely used to 
give insight into the high-frequency characteristics of a transistor amplifier. 
The hybrid-7r, which is shown in Fig. 7.20, is a type of j-parameter 
circuit. The validity of this circuit will be investigated by comparing its 
parameters to those of the h parameters at low frequencies. 



268 



Electronic Engineering 



The input resistance of the hybrid-^ with output shorted is very nearly 
r b + (h u + X)r e since r c is very large compared with (h fe + \)r e . This 
input resistance is essentially h ie . The ratio of the open circuit input 
voltage to a voltage v ce applied to the output terminals is (h u + \)r e jr c — 
r J r e = Ke- The short circuit output current is g m v' be . This current must 
be equal to h u i b at low frequencies if the circuit is valid. Then 



But, from Fig. 7.20 



= h 



fe'b 



gmVl 

>e = hih, e + l)r e 



(7.73) 
(7.74) 



Bo- 



n 
■AMr 



(h/e+ l)r e > ^=Ci 



6' 
-o— 



C c 



-AA/V- 



gmV'l 



'f*(J) T= 



-oC 



■ Co- 



Then 



and 



Fig. 7.20. The hybrid-w equivalent circuit. 
gmhiK + \)r e = h u i b 



on 



l fe 



(h fe + IK 



(7.75) 



(7.76) 



But, as shown in Chapter 4, 1/r, = g e = ^- \J Bl 
a. 



Iw.= 



A/. + 1 



Also since h te j(h u + 1) = 



(7.77) 



Then 



->cl 



(7.78) 



where I c is the g-point value of collector current. 

The capacitor Q was identified in the preceding section as the sum of 
the junction capacitance C e plus the diffusion capacitance C D . In Fig. 
7.18 of that section, Q was shown as being in parallel with r„ and in Fig. 
7.20, Q is shown as being in parallel with (h fe + \)r e . However, the 
circuit of Fig. 7.18 is a common-base circuit and the emitter current flows 



Small-Signal Amplifiers 



269 



through r e . In the common emitter circuit of Fig. 7.20, only the base 
current (approximately) flows through the fictitious resistance (h u + \)r e . 
The voltage developed across these two resistances is, therefore, the same 
in the two circuits. 

The capacitor C c represents the junction capacitance between the 
collector and the active portion of the base region. The capacitor C B 
represents the capacitance which exists between the collector and emitter 
terminals exclusive of C c . Thus C accounts primarily for header and 
lead capacitance plus the capacitance associated with the outer perimeter 
of the collector junction which does not involve r b . Manufacturers 
frequently list a value of common-base output capacitance with input 
open C ob . This value of C ob is essentially equal to C + C c . 



ib n 
b<^-VW- 



Cc 

-It- 



(h fe +l)r e < Cizt v 'be Q) < °=jc 



gm"'be 



<?C 



'Zl 



Co' 



E 



Fig. 7.21. A medium and high-frequency hybrid-" 1 circuit. 



At moderate and high frequencies, the reactance of C c is small in 
comparison with r c and the hybrid-^ circuit can be simplified as shown in 
Fig. 7.21. In this circuit the admittance g has been added in the output 
to account for the effect which base width modulation (represented by r c ) 
has on the output admittance. This g is a function of driving source 
resistance and is equal to h oe when the input is open (infinite source 
resistance). 

A frequency known as the beta cutoff frequency is frequently associated 
with a transistor. This is the frequency at which the short circuit forward 
current gain is reduced to 0.707 of the low-frequency value. Since g m is not 
a function of frequency, beta cutoff frequency occurs when v' be is at 0.707 
of its low-frequency value. But, assuming i b is constant, Fig. 7.21 shows 
that v' be will decrease to 0.707 of its low-frequency value when the react- 
ance of C x + C c in parallel is equal to (h u + l)r e . Therefore, 



1 



ft>« 



P (h u + l)r e (Q + C c ) h u + 1 
PROB. 7.15. Verify both forms of Eq. 7.79 assuming C x » C c . 



(7.79) 



270 



Electronic Engineering 



Bo— — WV 




Fig. 7.22. Equivalent circuit showing the effect of Z L on the effective shunt capacitance. 

This <Dp is the radian frequency at which p (or h fe ) drops to 0.707 P 
where /?„ is the low-frequency value of /3. 

When the transistor is used in an amplifier circuit, the base-collector 
junction capacitance is effectively multiplied by the voltage gain v ce jv' be 
as discussed in Section 7.4. As seen in Fig. 7.21, the current through the 
capacitance C c is proportional to the difference between v' be and v ce . 
Since v ce = —g m v' be Z L , the circuit of Fig. 7.22 is essentially equivalent 
to the circuit of Fig. 7.21. The capacitance C' has been replaced by C in 
Fig. 7.22. This capacitance C accounts for the effect which C c has on the 
output capacitance as well as the effect of C . The current through C c 
affects the voltage v' be . Thus, C is a function of the driving source 
resistance as well as the frequency and is usually much larger than C ob . 
For complete equivalence, a conductance G = C c (imag g m Z L ) should 
be placed in parallel with C x . This conductance is usually negligible in 
R-C coupled amplifiers, but becomes important in tuned amplifiers 
(Chapter 8). 

In the preceding discussion, the transistor was assumed to be driven by 
a current source /„. A comparison of Fig. 7.22 and Fig. 7.21 will show 
that the upper cutoff frequency of the amplifier will decrease from f fi 
when the real part of the load impedance increases from zero, assuming 




d + C c (l +R e \g m Z L \) 



Fig. 7.23. Equivalent circuit showing the effect of source resistance R s on the frequency 
response. 



Small-Signal Amplifiers 271 

/„ to be constant. In contrast, the cutoff frequency of the amplifier may 
increase fromf p if the transistor is driven from a source which has finite 
resistance R s , as illustrated in Fig. 7.23. The resistance {h fe + l)r e is in 
parallel with r b + R s , and thus the shunt resistance is reduced. The 
upper cutoff frequency in radians per second is 

1 1 (7.80) 



RshCsk . P,. + IK I (r b + RMCt + C c (l + Re[g m Z L ])] 
where the parallel bars | indicate the parallel combination of (h u + l)r e 
and (r b + R s ). From Eq. 7.80, it can be seen that the maximum upper 
cutoff frequency is attained when both source and load resistances 
approach zero. Then 

max co 2 = 7. (7.81) 

[(*/. + IK || r^C, + C c ] 

This maximum value of co 2 is sometimes called the transverse cutoff 
frequency. 

The value of w 2 calculated from Eq. 7.80 will be higher than the actual 
cutoff frequency because the stray wiring capacitance and other shunt 
capacitances external to the transistor on the input side have not been 
included in the circuit. The resistance r b prevents the inclusion of these 
capacitances directly in parallel with C x . However, in a high frequency 
transistor, r b is usually quite small and only a slight chance for error is 
introduced by lumping the input circuit capacitance with C x . In lower 
frequency transistors, the stray circuit capacitance is usually negligible 
in comparison with the diffusion capacitance. 

The junction capacitance of a field effect transistor can be included in 
the conventional y parameter circuit given in Chapter 6. However, the 
capacitance between the gate and the drain has an effectiveness 

CJ\ + R e \g m Z L \) 

as previously discussed. Therefore, the high-frequency equivalent circuit 
for a field effect transistor is as shown in Fig. 7.24. The total shunt 
capacitance is 

C sh = C W + C gs + C gd (\ + R e \g m Z L \) (7.82) 

where 

C w is the stray circuit capacitance 

C gs is the gate-source capacitance 

Then the upper cutoff frequency in radians per second is 

1 



Rsh^sh 



(7.83) 



272 



Electronic Engineering 



i.(T) B. 



G 

— o- 



Cg.+ C gd (l+Re[g m Z I .)) 
/ 

gmVgs 




Fig. 7.24. High-frequency equivalent circuit for a field-effect transistor. 

where R sh is the parallel combination of R g and the driving source re- 
sistance R gen . Since the FET is especially attractive as an amplifier for 
signals which are provided by a high impedance source, some compromise 
of voltage gain may be necessary in applications which require a high 
value of eo 2 . The term C gd {\ + Re \g m Z L \) is usually the predominant 
circuit capacitance. 

PROB. 7.16. A given transistor has h u = 100, h te = 1750 n, h oe = 50,umhos, 
h re = 6.25 x 10 -4 , C o6 = 10 pf and w fi = 4 x 10* radians per second at the 
quiescent operating point Ic — 2.0 ma and Vcb = 5 volts. Assuming the 
capacitance C to be 1 pf, determine the value of the elements of the hybrid w 
circuit shown in Fig. 7.20, including g m , at the operating point given. Assume 
T = 17°C. 

PROB. 7.17. A high-frequency transistor which is operated at I c = 10 ma and 
V CE = 10 volts has h u = 100, r b = 100 Q, f = 3.0 mc, and C c = 6 pf. The 
a-c load resistance is 500/0° ii and the driving source resistance is 1 Kfi. 

a. Determine the upper cutoff frequency of the amplifier. 

b. Determine the upper cutoff frequency if the driving source resistance is 
reduced to 100 ohms. 






7.8 GAIN-BANDWIDTH PRODUCT 

The product of gain and bandwidth is essentially constant for a specific 
tube. The bandwidth B = co 2 — co u but co 1 is usually very small compared 
with co 2 . Therefore, B c^ eo 2 . The gain is considered equal to the reference 
gain throughout the useful range of the amplifier. Therefore, the gain- 
bandwidth product for a tube amplifier is 



But 



then 



GB = g m R s 



K-sh.t-'&i 



(7.84) 



(7.85) 



GB = £2. 



(7.86) 



Small-Signal Amplifiers 273 

When several cascaded amplifiers are required for a specific application, 
the tube should be selected at least partially on the basis of the ratio of 
transconductance to total tube capacitance. This ratio is defined as the 
figure of merit for a tube. The figure of merit of a stage could be defined 
as the ratio of the transconductance of the tube to the total capacitance 
of the stage C sh . 
Table 7.1 lists the figure of merit for several tubes. Triodes are not 

TABLE 7.1 
Figure of Merit for Some Typical Pentodes 

Tube Q C gp C g m gmlCsh 

Type /U/uf fi/if /ipf ,umhos 10 2 radians/sec 



6CB6 


6.5 


0.010 


3.0 


6,200 


653 


6AK5 


4.0 


0.020 


2.8 


4,300 


630 


6AC7 


11.0 


0.015 


5.0 


9,000 


562 


6AU6 


5.5 


0.003 


5.0 


5,200 


495 


6SJ7 


6.0 


0.003 


7.0 


2,000 


154 



included because their effective capacitance is dependent on the gain of 
the amplifier. 

The figure of merit of a transistor is usually considered to be the 
product of low-frequency common emitter current gain and the beta 
cutoff frequency. Thus 

where /, is the frequency at which the extrapolated value of common 
emitter current gain is unity. The alpha cutoff frequency is approximately 
equal to/ r . 

7.9 TRANSFORMER COUPLING 

The transformer has some inherent advantages as a coupling device. 
For example, the ohmic resistance of the windings may be small in com- 
parison with the a-c impedance. As a result, the efficiency may be much 
higher than in the R-C coupled amplifier. In addition, the turns ratio may 
be chosen so that impedance matching may be utilized to obtain maximum 
power gain, and hence, maximum voltage or current gain. For example, 
the input impedance of a vacuum tube may be higher than its output 
impedance; therefore, a step-up transformer could be used to increase 
the voltage amplification. The input impedance in a transistor is usually 
lower than the output impedance. Consequently, a step-down transformer 



274 



Electronic Engineering 



could be used to increase the current gain of the amplifier. In each case 
the gain is approximately equal to the product of the gain of the amplifying 
device and the turns ratio of the transformer. 

The circuit diagram of a typical transformer coupled transistor amplifier 
is shown in Fig. 7.25. Stabilized bias is used for this example. Observe, 
however, that signal currents do not flow through the bias resistors in the 
circuit shown. Therefore, a high degree of temperature stability may be 
realized without sacrificing signal power gain. In this circuit, the ohmic 




Fig. 7.25. A transformer coupled transistor amplifier. 

resistance of the transformer secondary is part of the equivalent bias resist- 
ance R B . The design of the amplifier may be carried out with the aid of the 
characteristic curves provided in Fig. 7.26. As a first step, the collector 
supply voltage V cc is determined by the power output requirements of the 
amplifier. In this example, 9 v will be used for V cc . The d-c load resistance 
is the ohmic resistance of the primary winding, which is normally quite 
small. Consequently, the d-c load line is almost vertical, as illustrated 
in Fig. 7.26. To obtain maximum power gain and hence maximum current 
gain, the a-c load line is drawn with a slope approximately equal to that 
of the collector characteristic at the operating point. 4 This small slope 
permits the quiescent point to be chosen at a low value of collector current. 
For convenience, the point selected in the illustration is the intersection of 
the d-c load line with the curve i B = — 30 pa.. The relatively small collector 
current coupled with the small d-c voltage drop in the transformer primary 
yields a collector circuit efficiency which is high compared with that of an 
R-C coupled amplifier. 

4 Although the output admittance of the transistor may differ appreciably from h , 
the impedance match will be reasonably good if the output resistance is assumed to be 
l/h . A more precise matching procedure is presented in Chapter 9. 



Small-Signal Amplifiers 



275 



After the quiescent point has been selected, the bias circuit components 
may be determined in accordance with the temperature stability require- 
ments as discussed in Chapter 5. 

An equivalent circuit of the transformer coupled amplifier is given in 
Fig. 7.27. This equivalent circuit may represent either a transistor or a 
tube amplifier, depending on the specification of the current /. At this time, 
the transformer will be assumed ideal. That is, the coefficient of coupling 




-20 -25 

Vce in volts 



-40 



Fig. 7.26. Characteristic curves used for the design of a transformer coupled transistor 
amplifier. 

will be considered unity, the magnetizing current negligible, and the ohmic 
resistance of each winding negligible. Then the voltage and current 
ratios of the transformer are proportional to the turns ratio. In Fig. 7.27 
the ratio of primary turns to secondary turns is a. Then the source current 

(7.88) 
(7.89) 
(7.90) 



_a£ 2 i 2 _££ 2 ,_£ 2 _ 
R a R aR f 



v n = -I 



a 2 ** + R„ 



Maximum power will be transferred to the load when R = a^. In 

this case . «„ «. . _ 

MaRf) _ ~ IaR < (7 91) 

2a 2 /?, 2 



v n =-I- 



276 



Electronic Engineering 



(V) R„> av 



Ri\ 



Fig. 7.27. An equivalent circuit for a transformer coupled amplifier. 



The secondary current 



_ % _ al 
'~R<~ 2 



(7.92) 



When the amplifier is a common emitter transistor, / = h u i b . Maximum 
power transfer is attained when 



In = 



ah fe i b 



Then 



K = Is = a Jhi = W^f 
1 i» 2 2 W 



(7.93) 
(7.94) 



Since R t is usually much smaller than R in the case of a transistor, a step- 
down transformer would be used (a > 1). Equation 7.94 shows that the 
current gain will exceed h u when a exceeds two. 

When the amplifier employs a vacuum tube / = g m v g and, when Eq. 
7.91 for maximum power gain is used, 



And 



K„ = 



g m v g aRi 



= Vo = g m aR { _ g m R 



(7.95) 



(7.96) 



v„ 2 2a 

since R — a 2 R ( . 

For a triode tube, R t is usually large in comparison with ^ which is r„. 
Then a step-up transformer is used (a < 1). When \ja exceeds 2, the 
voltage gain exceeds g m r P which is fi. 

PROB. 7.18. Calculate the current gain of the transistor amplifier of Fig. 7.26 if 
Ro — II he = 30 Kfi, Rt ~ h u = 1.5 Kfi, and h u = 50. Specify the turns ratio 
of the transformer. Compare this gain with that of an R-C coupled amplifier 
using the same transistors. Assume R and R t to be the same in each case and 
neglect the signal currents through the biasing resistors. Also assume that the 
d-c load resistor R c in the R-C coupled amplifier is large in comparison with R ( . 
Answer: a = 4.47, K t = 111.7, K t = 45.4 for R-C amp. 



Small-Signal Amplifiers 



277 



The approximation that R = \\h oe and R t = h ie will not be valid under 
all conditions. Accurate values of R t may be obtained from the equivalent 
circuit of a particular transistor configuration if the load impedance is 
known. Therefore, the design of an amplifier normally proceeds from the 
load toward the input. Since the transistor does not provide complete 
isolation of its input circuit from its output circuit, a simultaneous solution 
of the entire amplifier circuit would be required in order to obtain ideal 
impedance matching for a specified set of conditions. However, this 
method of solution may be very complex and tedious. Fortunately, the 
order of accuracy required does not justify a simultaneous solution. The 
output impedance of the transistor may be obtained with sufficient accuracy 
when the input is furnished by a transformer if it is asssumed that the 
driving source impedance is h u . 

7.10 FREQUENCY RESPONSE OF A TRANSFORMER 
COUPLED AMPLIFIER 

The transformer may be represented by an equivalent Tee as illustrated 
in Fig. 7.28a. 5 R x and L x are the resistance and leakage inductance of the 
primary. Also, the resistance and leakage inductance of the secondary, 
R 2 and L 2 , have been referred to the primary by the multiplication factor a 2 , 
where a is the turns ratio of the primary to secondary. In addition, the 
mutual inductance M and load resistance R t have been referred to the 
primary. The distributed winding capacitance and circuit wiring capaci- 
tance have been neglected. 

Basic text books which treat inductively coupled circuits show that the 



<P 






aM* 



3 



a 2 Ri' 



(a) 




h(£> R ^ L »% 



av2 



(b) 



Fig. 7.28. (a) The equivalent T circuit is used to represent the transformer in a trans- 
former coupled amplifier, (b) The low-frequency equivalent of (a) for a tightly coupled 
transformer. 

6 Hugh Hildredth Skilling, Electrical Engineering Circuits, Second Edition, Wiley, 
p. 341, Mew York, 1965. 



2 ?8 Electronic Engineering 

mutual inductance is 

M = k(LuL^ (7.97) 

where k is the coupling coefficient, Z, u is the self-inductance of the primary, 
and L 22 is the self-inductance of the secondary. Then, since L 22 = Z, n /a 2 , 

aM = kL n (7.98) 

Figure 7.28 shows that the total self-inductance of the primary is 

L u = L x + aM = L x + kL n 
Then 

A = (1 - k)L n (7.99) 

The term 1 — k is the fraction of primary flux which does not couple the 
secondary winding. Then L t is the leakage inductance of the primary. A 
typical coupling transformer has a coefficient of coupling k of the order of 
0.99 or higher. Therefore, at low and middle frequencies, the leakage 
inductances may be neglected because their reactances are small in 
comparison with the reactance of the primary and the associated amplifier 
resistances R and R ( . Also the resistance of the transformer windings, 
Ri and R^ are negligible in comparison with R and R t . Consequently, a 
simplified low-frequency circuit may be drawn as shown in Fig. 7.286, 
where R is the parallel combination of R and a 2 R { . The output voltage is 

aF2 = - /i ^ r iz: (7100) 

aF2= - 7i *;ri^ (7101) 

When maximum power gain is achieved, a*/?, = R and R = RJ2. Let 
«>! be the frequency at which the reactance of the primary is equal to the 
resistance R. Then 



R 

A U 

and 



(7.102) 



<*V% = ~hR (7.103) 

As for the R-C coupled amplifier, Eq. 7.103 can be reduced to 

G = -K— — (7.104) 

s + co x 

where K is the reference gain of the transformer coupled amplifier (Eq. 7.94 
or Eq. 7.96). The negative sign may or may not precede K, depending on 



Small-Signal Amplifiers 



279 



the method of connecting the transformer. The gain equation has the 
same form as the R-C coupled amplifier in the low-frequency range. The 
frequency, phase, and time response would be identical in the two types of 
amplifiers in this range, providing that a) x is the same in each case. 

PROB. 7.19. A common emitter transistor amplifier has R = 20 Kii. A cou- 
pling transformer provides an impedance match to the following transistor. What 
primary inductance will be required if/ x = 30 Hz? Answer: L = 53 h. 
PROB. 7.20. Approximately what primary inductance will be required for the 
transformer of Prob. 7.19 if a 10% sag is permitted for a rectangular pulse of 
1 msec? 

The chief disadvantage of the transformer as a coupling device may be 
the large inductance required for satisfactory amplification of long pulses 
or low frequencies. Under these conditions, the transformer would be 



CD 1 T C " 



■VWvV — r ^W K 



a*R 2 



a 2 L2 




AA/Wv — r olfo" >L - 



aM 



4= ><* 2 Ri 



Fig. 7.29. High-frequency equivalent circuit for a transformer-coupled amplifier. 



costly and bulky. As a result, the cost or weight may be a major concern 
and may outweigh the several advantages of a transformer. Miniature 
transformers are available for coupling transistor amplifiers. However, 
those designed for audio frequency amplification generally have poor low- 
frequency response. In contrast, rectangular pulses of rather short 
duration [a few microseconds] may be amplified with little distortion when 
a miniature transformer is employed. 

The leakage inductance cannot be neglected at high frequencies or large 
values of s. The shunt capacitance of the circuit, as well as distributed 
capacitance of the transformer, also needs to be considered. An equivalent 
circuit which includes the circuit capacitance is given in Fig. 7.29. In this 
circuit, the capacitance C includes the distributed capacitance of the 
primary, and C, includes the distributed capacitance of the secondary. As 
a word of caution, considerable inaccuracy results from representing 
distributed capacitance by a lumped capacitor. However, fairly good 
qualitative ideas may be gained concerning the transformer performance. 



280 



Electronic Engineering 



If a highly accurate equivalent circuit were devised, the circuit complexity 
would present a formidable solution. The reactance of the magnetizing 
inductance aM is so large at high frequencies that it can be removed from 
the equivalent circuit. Consequently, the leakage inductance a 2 L 2 can be 
combined with Z, x to produce a total leakage inductance 2Z, X . This follows 
because 

L 2 = (1 - k)L 22 (7.105) 

or 

a*L 2 = a\\ - k)L 22 (7.106) 

but since a 2 L 22 = L 1U 

a 2 L 2 = (1 - k)L n = L x (7.107) 

The simplified equivalent circuit is presented in Fig. 7.30a. A further 
simplification is shown in Fig. 7.306. In this circuit the secondary resistance 
referred to the primary a 2 R 2 has been lumped with the primary ohmic 




Rl 2Li a 2 R 2 

vW-^WiTWvV- 



o2 ■ 



-a 2 Ri 



fi-VW^RftP-? 



0) 



► i?„ 



=f=c. 



cr- 



\Ri 



Fig. 7.30. Simplified high-frequency equivalent circuits of the transformer-coupled 
amplifier. 

resistance to form R. To simplify the symbolism, symbol CJa 2 has been 
changed to C' t and a 2 R t has been changed to R' t . The nodal equations 
for the circuit of Fig. 7.306 are 



( c - +sC - + ;nhz)''-<-;rhz Pi --'] 

where 

G = \IR and G\ = l/R', 
Solving for V B , we find that 

-I/(R + sL) 



(7.108) 



V„ = 



(g + sC + — !— ) (g\ + sC'i + —1—) l - 

\ R + sL/\ R + sL/ (R + 



(R + sLf 



(7.109) 



Small-Signal Amplifiers 281 

Expanding the denominator, we have 

V B . -//(* + sL) (7.110) 

(G + sC )(G { + sC { ) -\ — — 

R + sL 

The transfer function of the transformer is 

_ V B _ 1 

° ~ 7 ~ " (G + sC )(G'i + sC' t W + sL) + (C„ + CJs + G + G',. 

(7.111) 
1 
C C T 

\ Cj\ C'J\ LI C C t L\ C + C'J 

Equation 7.112 is a third-order equation. 

This third-order equation can be solved by conventional means if the 
coefficients are in numerical form. However, if a general solution is 
required, all possible numerical answers must be found. The root-locus 
technique allows us to make a plot of all possible roots in the denominator. 
This root-locus method is briefly presented in Appendix II. 

Equation 7.112 may be arranged in proper form for root-locus solution 
as follows. 

1 



G = 



^4+t)(^)K) 



i + c ° +c < 



C C\L 
Equation 7.113 is of the form 



s + (G„ + G',.)/(C„ + C\) 

LK)(-?;)K). 



(7.113) 



G = — - — (7.114) 

1 + HF 

where N is the numerator, H = (C„ + C \)I(C C \L) and F is the fraction 
in the denominator which involves s. 

In applying the root-locus technique, the poles and zeros of F are first 
plotted on the s plane as shown in Fig. 7.31 . Then the locus of all values of 
s which cause the angle of F to be 180° is drawn as indicated in Fig. 7.31. 
Again, the object is to factor the denominator of Eq. 7.112. The roots of 
the denominator occur at values of s which make HF = l /l80° . Conse- 
quently, a value of s must be found for which H = l/\F\ on each branch 



282 



A 

-X-* — o- 



Go &2 G + G'j G'j 
Co Co + C'i Ci 




Electronic Engineering 

joi 



Fig. 7.31. The root-locus plot of the denominator of Eq. 7.113. 



of the locus. The spirule, which is a device especially designed for the 
solution of root-locus problems, may be used to locate the point for which 
H = (C + C'JKCoC'tL) on each branch of the locus. 

Each root-locus branch begins on a pole of F and terminates on a zero 
of F, either finite or at infinity. Thus, there are as many root-locus 
branches as there are poles of F. In this figure, there are three branches. 
Since the value of H must be zero at a pole of F and infinite at a zero of F, 
the values of s which are the roots of the denominator of Eq. 7.112 move 
away from the poles of F as H increases. Therefore, when H is small, all 
the roots lie on the negative real axis. The time response, then, which is 
obtained by taking the inverse transform of Eq. 7.112, would consist of 
exponential functions. These functions are of the same form as those 
obtained for the R-C coupled amplifier and result in the same general type 
of response as the R-C coupled amplifier. In order for H to be small, 
however, the leakage inductance L and the effective winding capacitances 
C„ and C' ( must be large. But C , C\, and L, in conjunction with their 
associated resistances (or conductances), determine the location of the 
poles of F. To have good high-frequency response or short rise times, the 
poles of the transfer function and hence the poles of F must occur at large 
values of s. Then C , C' f , and L should each be small to provide good 
frequency and time response. This response criterion usually leads to values 
of H which place the poles of the transfer function (Eq. 7.1 13) off the real 
axis on the two branches which depart from the real axis and seek zeros at 



Small-Signal Amplifiers 



283 



infinity. A typical location of the poles of the transfer function might be 
at the points labeled A in Fig. 7.31. The response here is said to be under- 
damped in contrast to the overdamped case in which all the poles are 
located on the real axis. Critical damping occurs when a double pole is 
located at the point where the root-locus branches depart from the real 



axis. 



In the usual underdamped case, the transfer function of the transformer- 
coupled amplifier may be written as follows. 



llC C' t L 



/-* — K 

(s + Oj + jco )(s + a t — ja) )(s + <r 2 ) 

where a x , co , and <x 2 are obtained from Fig. 7.31 

G--K llC ° CiL 

(S + CT 2 )(S 2 + 2CTjS + ffx 2 + co 2 ) 

Equation 7.116 may be written in the form 

1/C.CL 



G= -K 



(s + o 2 )(s 2 + 2£o>.s + w „ 2 ) 



(7.115) 



(7.116) 



(7.117) 



where co n is the natural resonant frequency (a>„ 2 + a^f A and £ is the 
damping ratio = o x \(o n . 

The pole-zero plot of a typical transformer-coupled amplifier, including 
the low frequency pole and zero, is shown in Fig. 7.32. Since the damping 
ratio £ = <Ti/co„, the cosine of the angle d is also equal to £. The complete 
transfer function of the transformer-coupled amplifier can be written from 




Fig. 7.32. The complete pole-zero plot of a typical transformer-coupled amplifier. 



284 



Electronic Engineering 



f = 0.8 




2 4 6 8 10 12 
Time in microseconds >- 



Fig. 7.33. Front edge response of a transformer-coupled amplifier, 
inspection of Eqs. 7.104, 7.117, and Fig. 7.32. 



G = -K 



slC e C t L 



(s + WiXs + ff 2 )(s 2 + 2£a>„s + a>„ 2 ) 



(7.118) 



The factors £ and o> n primarily determine the high-frequency or rise 
time response characteristics of the transformer. The front edge or rise 
response to a step input voltage is plotted in Fig. 7.33a for various values 
of £, with co„ held constant. The response is plotted for various values of 
<i)„ in Fig. 7.336 with £ held constant. The frequency response of the 
amplifier is presented in Fig. 7.34 for various values of £. It is evident 
from the figures that a high resonant frequency, m n , will provide a short 
rise time and good high-frequency response. The high value of <w„ can be 
accomplished by providing small values of C , C' and L. For a specific 
value of coupling, the leakage inductance is proportional to the primary 
inductance which is set by the value of R and the desired co u as previously 
discussed. For a specific value of primary inductance the leakage induct- 
ance can be reduced by increasing the coefficient of coupling k. High 
permeability cores and special winding techniques are frequently employed 
in high-quality transformers to provide very tight coupling. 




Fig. 7.34. Frequency response of a transformer-coupled amplifier. 



Small-Signal Amplifiers 285 

The distributed capacitance of the windings increases with the number 
of turns. Consequently, a step-up transformer would have a comparatively 
large secondary capacitance which is divided by a 2 in referring it to the 
primary. Thus C' t may become very large if a large step-up turns ratio is 
used. As previously discussed, w„ decreases as C't increases. As a 
consequence, the step-up ratio of the transformer is restricted by the 
acceptable value of resonant frequency. Usually, a step-up ratio of 3 is 
about the maximum usable value for a good quality transformer. In 
contrast, C\ may be very small when a step-down transformer is used. 
Reducing the reluctance of the magnetic path is helpful in reducing winding 
capacitance, because the number of turns would be reduced for a specific 
value of inductance. In addition, special winding techniques may also be 
used to minimize the distributed capacitance. The transformer connec- 
tions recommended by the manufacturer must be used to realize minimum 
winding capacitance. 

From Figs. 7.33 and 7.34 it is evident that a value of £ much lower than 
0.8 would cause an appreciable overshoot when the input is a step function, 
or an appreciable peak in the steady-state frequency response. Of course, 
small values of leakage inductance are helpful in producing large values 
of £. In addition, the resistance of the windings may be used as a control 
parameter. However, an increased winding resistance may result in a 
reduced efficiency. 

7.11 THE COMMON COLLECTOR CONFIGURATION 

In the preceding paragraphs, some of the shortcomings of the transformer 
have been discussed. From this discussion, it is evident that the bandwidth 
of the transformer is limited by its ratio of primary inductance to leakage 
inductance. As a consequence, a device which would perform the function 
of impedance transformation throughout a wider range of frequencies 
would be very useful. Fortunately, the common collector amplifier 
configuration meets these requirements in addition to increasing the power 
level of the signal. The v acuum tube version of this type of amplifier is 
commonly known as a cathode follower . Similarly, the transistor version 
is frequently known as an emitter follower . 

Figure 7.35a is the circuit diagram of a typical common collector 
transistor amplifier. In this circuit, the load is placed between the emitter 
and the common ground terminal. Furthermoie, the input signal is 
applied between the base and the common ground terminal. The collector 
is connected directly to the collector supply voltage V cc . Since the collec- 
tor supply must be adequately bypassed, the collector is maintained at 
the same signal or a-c potential as the common ground terminal. Thus the 



286 



Electronic Engineering 




v cc £ — WWV- j— vWW f 




r m ib 



<b) 

Fig. 7.35. The common collector transistor amplifier, (a) Circuit diagram; (b) equiv- 
alent circuit. 



time-varying input signal actually appears between the base and the 
collector, whereas the output signal appears between the emitter and collec- 
tor, hence the name common collector or grounded collector. 

An equivalent Tee circuit of the common collector connection is shown 
in Fig. 7.356. In this circuit, the short-circuit current gain may be 
determined by letting R L = 0. Then 



— h = h + i c = h + h/eh 
K = -= -(1 + h, e ) 



(7.119) 
(7.120) 



Equation 7.120 shows that the forward current amplification factor of the 
common collector configuration is slightly greater in magnitude than that 
of the common emitter connection. The input voltage v t is found from 
the equation 

»< = »» + »* (7.121) 



-^ 














i p 


hi c <. 








1 + 


(J 


hf c ib t 


>h x V2 


O 1 






i 




(ft 



loe V2 



I 



(a) (b) 

Fig. 7.36. The A-parameter equivalent circuit for the common collector configuration 
(a) using common collector h parameters and (b) using common emitter h parameters. 



Small-Signal Amplifiers 287 

The base-to-emitter voltage v be is normally small compared with v . 
Therefore, the voltage amplification is approximately unity, and the power 
gain is approximately equal to the current gain. Therefore, the power gain 
of the common collector is smaller than that of the common emitter 
configuration. 

The input and output impedances of the common collector configuration 
may be determined in terms of the equivalent Tee parameters by writing 
Kirchhoff's equations for the circuit of Fig. 7.35Z>. However, it is more 
convenient to use an /(-parameter equivalent circuit as shown in Fig. 7.36a. 
The forward current amplification factor h fe has already been found in 
terms of the common emitter current amplification factor (Eq. 7.120). 
Thus 

-h fc = h re + 1 ~ h u (7.122) 

Figure 7.356 shows that with the output shorted, the common collector 
configuration is indistinguishable from the common emitter configuration. 
Therefore, the low-frequency input resistance with the output shorted is 

h ic = h ie (7.123) 

When the input is open, it can be seen from Fig. 7.356 that the output 
resistance is r e + r d ,-since i b , and hence r m i b , is equal to zero. Then 

h oc = h oe (7.124) 

Figure 7.356 also shows that 

h rc = — r -&- ~ 1 (7.125) 

Thus it should be observed that the common collector h parameters are 
almost identical in magnitude with the common emitter h parameters, 
with the exception of h rc which is practically unity. The common collector 
A-parameter equivalent circuit is redrawn in Fig. 7.36b, using the common 
emitter h parameters. 

With a load resistance R L placed in the emitter circuit the low-frequency 
input resistance may be calculated from the equivalent circuit of Fig. 
7.37a. Observing that there is no polarity reversal in the emitter follower, 
we see that 

v t = h ie i b + v 2 (7.126) 

but 

v 2 = {hfe + 1)ib (7.127) 

Ke + Gl 

where G L = l/R L . 



288 



Electronic Engineering 




(a) 



<b) 



Fig. 7.37. Equivalent circuits for calculating (a) input resistance and (6) output resistance 
of the common collector circuit. 



Substituting Eq. 7.127 into Eq. 7.126, we have 



»i = Kh + 



The input resistance 



Ke + G L 

+ 1 



h Ke. + G L 



Usually G L is large in comparison with h oe . In this case, 
R t ^ K + (h f . + l)R L 



(7.128) 



(7.129) 



(7.130) 



PROB. 7.21. Show that the input resistance of the emitter follower approaches 
r c as the load resistance becomes very large. 

The equivalent circuit of Fig. 7.376 may be used to obtain the mid- 
frequency output conductance when the driving source resistance is R g . 



Then 



'2 = KeV* - (h, e + l)i» 

hi. + R g 
(h fe + 1)^2 



-i* = 



h = K e v 2 + 



h ie + R g 

+ 1 



y = i* = h + ^- 

V 2 h ie + R„ 



(7.131) 
(7.132) 

(7.133) 
(7.134) 



Observe from Eq. 7.134 and Eq. 7.129 that the emitter follower is an 
impedance transformer with an impedance ratio equal to h u + 1. The 
total resistance in the output circuit is multiplied by h fe + 1 and added to 



Small-Signal Amplifiers 



289 



h ie to produce a very high input resistance for normal values of load 
resistance. Similarly, the total output conductance is h M plus the input 
circuit conductance multiplied by h fe + 1. 

PROB. 7.22. A transistor has the following coefficients: r e = 25 Cl,r b = 500 £1, 
r c = 1 megohm, a = 0.99. Calculate the input resistance of the common 
collector connection with (a) R L = 1 KQ. (b) R L = 20 Kii. Answer: (a) #< = 
94KCI. 

PROB. 7.23. Calculate the output resistance of the common collector configura- 
tion of the transistor above with driving source resistance R„ (a) 1 KQ, (b) 50 Kii. 




DO 

£ 

To 
o 
o. 



1M 




1 1 1 


CB 


^~ 


100 K 






S — 








—~~+*^CE 


10 K 


- 




/^ 


IK 






/ - 


100 




cc 


A \ r "~ 




^ re '*A.+ l 


10 


— 






1 




1 1 1 


1 1 1 



Td 



eo oj »h t-« 

I I I I 

o o o o 

1— I F-^ »-* «-H 



-Bft 
Fig. 7.38. Input and output impedance of the three transistor configurations. 

The variation of input impedance as a function of load resistance for 
each transistor configuration is plotted in Fig. 7.38a. Similarly, the 
output impedance is plotted as a function of driving source impedance in 
Fig. 7.3Sb. These impedances will all be resistive at mid-frequencies but 
will be complex in the low and high frequency ranges. As the frequency 
approaches beta cutoff (f p ), the current gain of the common collector 
configuration decreases in the same manner as in the common emitter 
configuration. In addition, the input impedance decreases in the same 
fashion as the current gain. 

The load lines and quiescent operating point of the common collector 
amplifier may be located on the collector characteristics in the same manner 
as for the common emitter configuration. The biasing circuits may also 
be similar to those of the common emitter. The d-c resistance of the load 
may be sufficient to provide good current stability with the stabilizing 
resistor (R 2 ) between the base and ground removed. The bias resistor 
(R 3 ) would then carry only the base bias current. This arrangement would 



290 



Electronic Engineering 





CE 


^^ 1 
1 




t 




1 


JS-i-O^ 


G P 


// ' 

// ^*^\ 

(^ li 


1 
1 
1 

l 


1 



r e 



Td 



Rl- 



Fig. 7.39. Relative power gain as a function of load resistance for the three transistor 
configurations. 

provide maximum input impedance which may be highly desirable. When 
the d-c resistance of the load does not provide adequate stability, an 
additional resistor, properly bypassed, could be placed in series with the 
load. 

Figure 7.39 shows the power gain as a function of load resistance for 
the three transistor configurations. The common emitter configuration 
always provides the maximum power gain, but it is approached by the 
common collector at small values of load resistance and by the common 
base at large values of load resistance. 

7.12 THE CATHODE FOLLOWER 

The v acuum tube version of the common collector amplifier is called 
the cathode follower. In this circuit, the load is in the cathode side bfthe 
circuit and the plate is connected directly to the V PP supply voltage as 



-oV f 




Fig. 7.40. Cathode-follower amplifiers. 



Small-Signal Amplifiers 



291 



illustrated in Fig. 7.40. Cathode or self-bias may be used in a fashion very 
similar to that of a common cathode amplifier as shown in Fig. 7.40a. 
The voltage drop across resistor .Rj is used to provide the desired d-c 
potential between cathode and grid. In this circuit the capacitor C is used 
to bypass the a-c components of the cathode current. On the other hand, 
the bias resistance /? x may be part of the load resistance Z L as shown in 
Fig. 7.40ft. The capacitor C is then unnecessary. In some circuits where a 
large amount of bias or a small load resistance is desired, the resistance R t 
may be the entire d-c load resistance. The grid resistor R g will then return 
to ground as in the common cathode amplifier. 




Fig. 7.41. An equivalent circuit for the cathode follower. 

Inspection of the circuits of Fig. 7.40 will reveal that for mid-range 
signals, 

v, = v gk + v a (7.135) 

A mid-frequency equivalent circuit of the cathode follower is provided 
in Fig. 7.41 (remember that /u is a negative term). Writing KirchhoiT's 
voltage equation around the loop, we have 



W* = i P ( r p + z l) 



i„ = 



" o = '„Zi = 



r v + Z L 

-fVgk Z L 



(7.136) 
(7.137) 

(7.138) 



Equation 7.138 would be more useful if the output voltage were obtained 
in terms of the input voltage v t instead of the grid to cathode voltage v gk . 
Using Eq. 7.135, we find that 

• v gk = v i -v (7.139) 



292 

Then 



tf„ = 



Electronic Engineering 

-r*z L (v t - v ) 



\ r„ + Z T I r 






r„ + Z. 
v (r B + Z L - fiZ L ) = -!*Z L v ( 

The reference gain 



G = -° = 



-1*2 1 



r P + (-/* + V)Z L 



(7.140) 

(7.141) 
(7.142) 

(7.143) 



or 



(7.144) 



c _ 1-rK-ft + Wi. 

[rji-p + 1)] + z L 

Equation 7.144 has the same form as the gain equation of the common 

cathode amplifier. However, both the amplification factor and plate 



f*p 



M>V 




H 



Fig. 7.42. Circuit for determining the output impedance of the cathode follower. 

resistance have been reduced by the factor (— /x + 1). This reduction 
causes the gain to approach —^/(—/j. + 1), as the load impedance becomes 
large in comparison with rj{—/x + 1). It is evident from this and from 
Eq. 7.135 that the voltage gain must be less than unity. 

To check the validity of the assumption that the output impedance of 
the amplifier is r v /(—fi + 1), a voltage v 2 is applied at the output terminals 
as shown in Fig. 7.42. Writing Kirchhoff 's voltage equation around the 
loop, we have 

»2 - I*"*, = i P r p (7.145) 

In case the grid resistor returns to ground, the grid remains at ground 
potential and "* 

v, = v v „ (7.146) 

(7.147) 



Then 



'kg 



f 2 (-/" + 1) = '>: 



»' jj 



Small-Signal Amplifiers 

and 



293 



t>2 

R„ = - 



-li + 1 



(7.148) 



Consequently, the equivalent circuits of Fig. 7.43 can be drawn for the 
cathode follower. 

In case the grid resistor is returned to a tap on the cathode circuit as 
shown in Fig. 7.40a, the grid does not remain at ground potential when a 
voltage i> 2 is applied to the output terminals; then, the output impedance 
rises. If the driving source impedance were infinite, the circuit of Fig. 7.40a 
would have an output impedance equal to r„ because v ik = 0. In practical 
cases, however, the grid resistance R g may be made large in comparison 

ip 





> r p 




J-M + l 




1 + 




J-n + i 


— — o 







BrnPi 



-H+l 



Fig. 7.43. A-c equivalent circuits for the cathode follower. 

with the driving source resistance and the equivalent circuits of Fig. 7.43 
are approximately correct. 

PROB. 7.24. Determine the output impedance and maximum obtainable gain 

from a 12AT7 tube (one triode) when used as a cathode follower. Assume that 

none of the output voltage is fed back into the grid circuit. Also assume p = 

-60 and r„ = 15 Kfi. Answer: Z = 246 H. 

PROB. 7.25. Develop an expression for the output impedance of a cathode 

follower when a fraction of the output voltage A; = r,\{r g + R a ) appears between 

grid and ground in the circuit of Fig. 7.40b. Let r g be the impedance of the 

driving source. 

PROB. 7.26. Calculate the voltage gain of a 12AT7 cathode follower when 

R L =5 KQ. Answer: G v = 0.938. 

The input admittance of the amplifier may be determined by applying 
an input voltage v t and solving for the resulting current. Referring to 
Fig. 7.44, we find that 



I, = sC„V t + sC ak {V i - V ) 



(7.149) 



294 Electronic Engineering 

Using the relationship V = GV t , we have 

h = V t [sC„ + sC ak (l - G)] 



(7.150) 
(7.151) 



Y a = -* = s[C gp + (1 - G)C gk ] 

and the effective input capacitance is 

C e ti = C„ + (1 - G)C a , (7.152) 

Since the gain of the cathode follower approaches unity and has a positive 
sign, the effect of C gk is negligibly small. If a pentode tube were used as 




oVpp 



°v PI 




Fig. 7.44. Circuit for determining input impedance of a cathode follower. 

illustrated in Fig. 7.44b, the grid-to-cathode capacitance would be essen- 
tially eliminated and the total effective input capacitance would be 
negligibly small. 

Referring again to Fig. 7.44a, we see that if R„ is much greater than 
R 2 or R lt then 

i r - V -^^ (7.153) 



R. 



where 



A = 



R, 



Ri + R* 



R* 
Rj 



In case G is real, 



(1 ~ AG) 



R _ V i _ R ° 

' L 1 - AG 



(7.154) 
(7.155) 

(7.156) 



Small-Signal Amplifiers 295 

When R 2 = 0, A = 0, R { = R a \ but if R 2 approaches R L , A approaches 
unity and 

R { ~-£s- (7.157) 

1 - G 

Since the voltage gain of the cathode follower may be almost unity, the 
current gain and power gain may be almost equal to the ratio of the input 
resistance to the load impedance. This would be true when the — fi of the 
tube is 10 or higher and the load impedance is large in comparison with 
the output impedance. 

PROB. 7.27. Calculate the effective input capacitance of a cathode follower 
which has C gk = 4 pf, C gP = 4 pf and G = 0.90. Answer: Q = 4.4 pf. 

PROB. 7.28. Calculate the effective input capacitance of a pentode cathode 
follower which has C t = 10 pf, G = 0.90 and C flp = 0.003 pf. 
PROB. 7.29. Calculate the input resistance of a cathode follower which has 
R g = 0.5 megohm, R t = 1 Kii, R 2 = 10 Kft, /i = -20, and r v = 10 Kft. 
Answer: R t = 2.94 MCI. 

The gain bandwidth product of a cathode follower circuit is approxi- 
mately the same as a common cathode amplifier. However, the input 
signal level may be much higher than that of the common cathode 
connection, because 

v f = »,* + ». (7.158) 

fi = » rt + Gv t (7.159) 

r 4 (l - G) = v gk (7.160) 

v t = v ' k (7.161) 

(1-G) 

As the gain approaches unity, the input voltage may be very large in 
comparison with the grid to cathode or bias voltage. 

Figure 7.45 illustrates a method of graphical solution of the cathode 
follower. A 300-v V PP supply voltage has been chosen and a load line 
drawn for R L = 20 K. A bias of — 2 v has been selected as suitable for 
symmetrical input signals. The required bias resistance R x is 400 Q 
since I P = 5 ma. The remainder of the d-c load resistor jR 2 is then 19,600 O. 
The d-c drop across the tube is 195 v and the d-c drop across the load 
resistance is 105 v. The maximum peak-to-peak grid-to-cathode voltage 
swing is 6 v. This yields a peak-to-peak voltage swing across the load 
resistance of 300 — 110 = 190 v. Consequently, the peak-to-peak input 
voltage is 190 + 6 = 196 v, since the peak-to-peak grid voltage is 6 v. 
The voltage gain is xf§ = 0.974. If the a-c load impedance is appreciably 
different than the d-c value, an a-c load line should be drawn through the 
quiescent point. 



296 



Electronic Engineering 



16 

14 

S 12 

1 10 

i 8 

I 6 
E 4 

2 





1 


1 *> 
* / 


1 




1 


1 




- ^^ 


v 


P 




V 




- 




/i\^/ 








¥ 




— 


/ 1 / 




"V 




X 


- 


- y 


yr\ s 








'^ f 


— 


y^ 


1^^ 








XoS^ 





50 100 150 200 250 300 350 

Plate volts 

Fig. 7.45. Graphical solution of the cathode follower. 



The distortion of the cathode or emitter follower is usually very low 
because the output voltage is so nearly equal to the input voltage at all 
times. If the input voltage were sinusoidal, the output current would be 
almost sinusoidal. This would require that the grid-cathode or base- 
emitter voltage be nonsinusoidal, as may be seen from Fig. 7.45 where the 
peak grid voltage is only 2 v on the positive half cycle but is 4 v on the 
negative half cycle. This grid-cathode voltage waveform is sketched in 
Fig. 7.46. If the operating point is assumed to remain at V G = — 2 v and 
the peak-to-peak amplitude of the input voltage is the maximum permissible 
value of 196 v, the output voltage would have a peak amplitude on the 
positive half cycle of 98 — 2 = 96 v and a peak amplitude on the 
negative half cycle of 98 — 4 = 94 v. This 2-v amplitude difference 
between the two half cycles of the output voltage is only about 1 % of the 
peak-to-peak amplitude of the output voltage, and therefore, represents a 
very small percentage of distortion. 6 Actually, the lack of symmetry of 




Fig. 7.46. A sketch of the grid-cathode voltage waveform of the cathode follower when 
the excitation is a sinusoid of maximum permissible amplitude. 

* A quantitative discussion of distortion is given in Chapter 10. 



Small-Signal Amplifiers 



297 



the output signal slightly increases the average plate current, which in turn 
slightly increases the bias. This increased bias tends to improve the output 
waveform. 

Frequently the input signal is a positive pulse instead of a symmetrical 
signal. Maximum signal-handling capability is then provided if the 
quiescent bias is near plate current cutoff. Under these conditions, the 
grid may be maintained at d-c ground potential and the entire d-c load 
resistance used as bias. In the foregoing example, the operating or Q 
point would be located at approximately V a = — 6 v, and V P = 300 v. 
This would permit a peak pulse input voltage of 196 v and a pulse output 
voltage of 190 v. 

PROB. 7.30. Given a tube with g m of 5000 //mhos and an r v of 1 megohm, (a) 
Design an R-C coupled amplifier with a pass band from 30 Hz to 2 MHz. 
Assume wiring and interelectrode capacitance is 50 pf. Also assume the grid 
resistor to the following stage has a value of 1 megohm, (b) What is the mid- 
band gain of this amplifier ? 

PROB. 7.31. Two pentode tubes are to be connected in cascade. The following 
parameters are listed for the tubes : g m = 2000 /*mhos ; r v = 1 megohm. The 
maximum permissible grid leak resistor is 1 megohm, (a) Design an R-C 
coupled amplifier using the two tubes in cascade. The frequency response should 
be down 3 db at 60 Hz and at 200,000 Hz. (Note that thegain of each stage must 
not be down 3 db at the frequencies listed.) (b) What is the maximum voltage 
gain possible from this amplifier? 

PROB. 7.32. A cathode follower is connected as shown in Fig. 7.47. Note that 
the total load resistance R L = Ri + R 2 . This load resistance is to be set at 
15,000 il. The input signal t> { is sinusoidal. The magnitude of the input signal, 
and the magnitude of the bias resistor R t are each of such a value that the grid- 
cathode voltage swings from zero volts at one extreme to —4.0 v at the other 
extreme. The internal impedance of the source that delivers v t is lOKii; the 




=-150v 



=-150v 



Fig. 7.47. The circuit configuration for Prob. 7.32. 



298 Electronic Engineering 

reactance of the coupling capacitor is 1 Kii; whereas, R g has a resistance of 
1 megohm. Find the value of R x in ohms. Find the maximum, average, and 
minimum values of voltages at the five different points indicated on the diagram. 
(v it v g , v a , v k , and v„). Express all voltages in respect to' point a as a zero reference. 
PROB. 7.33. A 2N2712 transistor is to be used as an amplifier to provide 25 
milliwatts of power to a 500 CI capacitively-coupled load. The driving-source 
resistance is 2.0 KO and is capacitively coupled. A thermal stability factor 
Sj = 10 is adequate for this application. Design the amplifier. Determine all 
component values assuming the low frequency cutoff to be approximately 30 Hz. 
Determine the upper cutoff frequency and the rise time of the amplifier. Neglect 
wiring capacitance. 



8 



Small Signal Tuned 
Amplifiers 



The need frequently arises for an amplifier which will amplify only those 
frequencies which lie within a certain frequency range, or band. This type 
of amplifier is known as a tuned amplifier or band pass amplifier. Radio 
and television receivers, for example, use tuned amplifiers to select one 
radio signal from the many which are being broadcast. Several types of 
tuned amplifiers are discussed in this chapter. The gain and bandwidth of 
each type will be of interest. It will be assumed, in the discussion of tuned 
amplifiers, that the input signal is a modulated signal which has a bas ic 
frequency w M . In this case the amplifier should be tuned to to , and the 
required bandwidth will depend on the nature of the modulation (or 
variation) of the signal. The bandwidth requirements of interrupted 
carrier modulation are considered in this chapter. However, the 
band-width requirements of other types of modulation are necessarily 
delayed until later chapters where these other types of modulation are 
considered. 

299 



300 



Electronic Engineering 



8.1 SINGLE TUNED, CAPACTITVELY COUPLED AMPLIFIERS 

The R-C coupled amplifier may be transformed into a band pass 
amplifier by replacement of the collector or plate circuit resistor with an 
inductor, as shown in Fig. 8.1. A current source equivalent<circuit which 




Fig. 8.1. (a) Tuned pentode amplifier with inductance in the plate circuit. (6) Common- 
emitter amplifier with an inductance in the collector circuit. 



will represent either of the amplifiers of Fig. 8.1 is given in Fig. 8.2. In 
this circuit, C is the total shunt capacitance and R sh is the total shunt 
resistance of the amplifier and R s is the effective series resistance of 
the inductor. The bypass and coupling capacitors do not appear in the 
equivalent circuit because their reactances should be negligible at the 
operating frequency. 

The equivalent circuit of Fig. 8.2 may be simplified if the series combina- 
tion of L and R s is transformed into an equivalent parallel combination, 
as follows. Considering the real frequency domain, we find that 



%l = R s + J">L 



Y T .= — = 



1 



R s + jo)L 



(8.1) 
(8.2) 



Q> 



=FC 



Fig. 8.2. Equivalent circuit of the amplifiers of Fig. 8.1 . 



Small Signal Tuned Amplifiers 

Rationalizing, we have 



Y T . = 



R a — j(oL 



or 



Y r . = 



R 2 + a> 2 L 2 
R s coL 



— J 



301 



(8.3) 



(8.4) 



R 2 + co 2 L 2 J R 2 + co 2 L 2 

This is of the form Y = G + jB. a>L is usually large in comparison with 
R s . When <oL ^ \0R„ which is known as the high Q case, 



R s _ . J_ 
o) 2 L 2 ^ wL 



(8.5) 



This admittance could be produced by a resistance in parallel with an 





(a) (b) 

Fig. 8.3. Equivalent circuits for a tuned amplifier. 



inductive admittance. The resistance 



But 



1 co 2 I3 

R p = -^ 

G R s 

(oL 



Qo = 



R, 



where Q is the Q of the coil at the specified value of w. Then 

R P ~ Q b ojL 



(8.6) 
(8.7) 

(8.8) 



In the high Q case (Q > 10), the percentage variation of either Q or 
to is negligible within the usable frequency range of the amplifier, as will 
be shown later. The effective parallel resistance R p may then be considered 
to be constant. 

It is evident from Eq. 8.5 that the inductance of the equivalent parallel 
circuit is approximately equal to that of the series combination because 
cdL„ ~ 1/5 = caL. The equivalent parallel circuit is shown in Fig. 8.3a. 



302 < Electronic Engineering 

The equivalent parallel resistance R v may be combined with R sh to obtain 
the total shunt resistance R as illustrated in Fig. 8.36. Using the relation- 
ship of Eq. 8.8, we may define the circuit Q, sometimes called the loaded 
G, as 

Q = ~^- = Ro> C (8.9) 

a> L 

At the resonant frequency co , the total impedance is R because the 
impedance of the lossless resonant circuit is infinite. Then the output 
voltage is 

V = -IR (8.10) 

Referring to Fig. 8.3ft, we may obtain the j-domain 1 output voltage V 
as a function of the s-domain current source /. 

K = — = =^ (8.11) 

(!/R) + sC + (l/sL) s 2 LC + s(LlR) + 1 



v - ~ sLI 



= (c'): 



LC[s 2 + (l/RQs + (I/ LQ] \C I s 2 + (l/RQs + (1/LC) 
Equation 8.12 may be written in the standard form as 



(8.12) 



\ C / s + 24w n s + co n 

where w n is the undamped resonant frequency (LC)~ l/i and the damping 
ratio £ = l/2Rco n C. From Eq. 8.9 we can see that £ = l/2£? in the high 
Q case because co c^ co n . 

In case the amplifier is a vacuum tube amplifier, the current I = g m V g 
and 

G = - = - — ; (8.14) 

V„ C s 2 + 2tw n s + < 

Now, from a consideration of the amplifier response to a negative step 
function of magnitude V, the output voltage is 

VG s V 1 

y _ _ r u _ Sin* * (g Y5) 

s C 5 2 + 2£co rl s + w„ 2 

1 The capital I ox V with lower case subscripts will be used to denote currents and 
voltages in the frequency domain. This should not cause confusion because if the cur- 
rent or voltage is in the complex (s) frequency domain, the equations will contain 
the parameter j. In the real (Jco) frequency domain the equations will contain the 
parameter 10. 



Small Signal Tuned Amplifiers 



303 



The inverse transform of Eq. 8.15 can be most readily obtained by 
arranging this equation into the standard form 



V„ = 



g m V 



1 



K = 



c ( S + so 2 + co n \i - & 

g m y (Qq 

2 _L ,.. 2 



(O C (S + £(O n f + Oi'- 



(8.16) 
(8.17) 



where co = <w„(l — J 8 ) 1 ' 




Fig. 8.4. The time response of a typical tuned amplifier to a negative step-voltage input 
of magnitude V. 



Taking the inverse transform of Eq. 8.17, we have 



v = ^- e*** 'sin m t 



(8.18) 



This time domain output voltage is sketched in Fig. 8.4. Although a step 
input voltage may be an unusual type of excitation for a tuned amplifier, 
the resulting response is interesting and informative. This exponentially 
decaying sinusoid is the transient response and is frequently known as 
ringing. The time constant of the decay is 2RC in contrast to the usual 
time constant RC obtained when a circuit contains resistance and 
capacitance only. 

As previously discussed, the tuned amplifier is usually tuned to the 
desired excitation frequency. The excitation will then be assumed to be 
v g = Vsinojj, and the ^-domain excitation is Kco /(j 2 + co 2 ). Using 



304 Electronic Engineering 

Eq. 8.14, we have 

v _ Vo>o G _ _ 8 m y m s 

" s 2 + co* C s 2 + co 2 s i + 2lco n s + co n 2 

Rearrangement of Eq. 8.19 into the form of Eq. 8.17 gives 

v _ £mY <M (8 20) 

C (s 2 + coMs + & n f + *>„ 2 ] 
Using partial fraction expansion, we find that 

g m v 



V — — 



As + B Cs + D 1 

^"^ + '- - --J 2 + «d ( } 



C Ls 2 + a> 2 (s + £co 

The arbitrary constants A, B, C, and D may be evaluated by equating the 
numerators of Eqs. 8.21 and 8.20 (after the equations have been reduced 
to a common denominator). 

(As + B)[(s + £co n ) 2 + <o*] + (Cs + D)(s 2 + w 2 ) = <o s (8.22) 

Expanding, we have 

As 3 + 2Afr n s 2 + At, 2 m n 2 s + Am*s + Bs 2 + 2B& n s + BfroJ 

+ Boy* + Cs 3 + Cco 2 s + Ds 2 + Dm 2 = (o s (8.23) 

Equating the coefficients of equal powers of s gives 

A + C = (8.24) 

2A£,a> n + B + D = (8.25) 

A?to n * + Am* + 2BCco n + Cm* = co (8.26) 

Bl 2 <a 2 + Bm 2 + Dm 2 = (8.27) 

Some simplifications can be made by recognizing that £ 2 co„ 2 is very 
small in comparison with u> 2 when the circuit Q is 10 or higher as previ- 
ously assumed. Then £ ~ 1/20. In this case, from Eq. 8.27, D ~ —B. 
Then from Eq. 8.25, A ~ and from Eq. 8.24, C ~ 0. Substituting these 
values into Eq. 8.26, we have 

2B£ Wn ~co (8.28) 

B~-^- and £>~-.B~--^- (8.29) 

2£<u n 2£co„ 



Equation 8.21 then becomes 
g m V 



V„= - 



V + co„ (s + 4<uJ + w -/ 



2£co„C\s 2 + o>, 



Small Signal Tuned Amplifiers 



305 



Since 2£<o„ = IfRC and g m R = K, the reference gain, or gain at the 
resonant frequency, 



V = -Kv(-^ ^ ) 

V + eo 2 (s + Zo> n ? + Wo V 

Taking the inverse transform, we find that 

v = -KV(sin a> t — e"^ ""' sin co t) 
From the substitution of 2£a>„ = IjRC, 

v„ = -KV(\ - e-' /2iJC )sin co t 

A sketch of the response which is expressed by Eq. 8.33 is given in Fig. 
8.5a along with a sketch of the sinusoidal excitation. In addition, the 
response to an interrupted or pulse modulated carrier is given in Fig. 8.5fe. 
In this case the exponential decay of the output following the cessation of 
the input was deduced from the known transient response of the RLC 



IftTeeoPTtcaO 



(8.31) 



(8.32) 



(8.33) 





IMTfeftuPTiwJ 



-KV 



(a) (b) 

Fig. 8.5. A sketch of the response of a tuned amplifier to (a) sinusoidal excitation of 
frequency ca„, switched on at time = 0; (6) interrupted or pulse-modulated carrier 
excitation of frequency a> . 



306 Electronic Engineering 

circuit. The form of this transient response was obtained from the first 
example in this section in which the excitation was a step voltage (see 
Fig. 8.4). 

The desirability of a tuned amplifier, like other amplifiers, is measured 
by both the amplification and the preservation of the waveform of the 
excitation. As shown in Fig. 8.5, the tuned amplifier cannot faithfully 
follow instantaneous changes in the excitation amplitude. In other words, 
the tuned amplifier has a rise and decay time which is very similar to the 
R-C coupled amplifier counterpart. Two essential differences exist 
between the tuned amplifier and the untuned or video amplifier. First, in 
the case of the tuned amplifier, it is t he envelop e of t he output signal and 
not the individual cycle which rises and decays exponentially when th e 
sinusoidal excitation is instantly started or stopped. S econd, the time 
c onsta nt of the envelope rise or decay is 2RC in contrast to the tim e 
c onstant RC of the untuned or video amplifier. 

Again, as in the untuned amplifier, the tuned amplifier would ideally 
have a very high gain and a very small rise time. But the reference gain and 
the envelope rise time are both proportional to the shunt resistance R. 
Therefore, these two criteria are in conflict. A figure of merit F a for the 
amplifier might be 

p _ reference gain ^ 

envelope rise time 

In the R-C coupled amplifier, the rise time (10%-90%) was found to be 
2.2RC. This rise time was discussed in Chapter 7. Then, since the time 
constant of the envelope rise is 2RC, or double that of the R-C amplifier, 
the 10%-90% envelope rise time is 4 ARC. Using this rise time, we can 
find that the figure of merit of the tuned vacuum tube amplifier is 

F. = -£=£- = -£=- (8.35) 

AARC 4AC 

Equation 8.35 shows that the ratio of gain to rise time improves as the 
shunt capacitance C decreases. Therefore, the maximum gain-rise time 
ratio would be obtained when only the stray circuit capacitance is used to 
tune the circuit. However, this stray capacitance usually varies appreci- 
ably with temperature, amplifier gain, or supply voltage (for a transistor). 
Thus the resonant frequency of the amplifier may drift appreciably unless 
some fixed, stable capacitance is included in the tuning capacitance. 

PROB. 8.1. A pentode tube which hasg m = 2000 /tmhos and r v = 1 megohm 
is used as an r-f amplifier. The total shunt capacitance in the plate circuit is 
25 pf. The grid resistor in the following amplifier is 1 megohm, 
(a) Calculate the inductance required to tune the plate circuit to 2 MHz. 



Small Signal Tuned Amplifiers 3 07 

(£) Calculate the reference voltage gain of the circuit if the Q of the coil is 100. 

(c) What is the circuit Q? Answer: (a) 253 fihen, (b) 388, (c) 62. 
PROB. 8.2. An input signal v, = 2 sin (4tt x 10 6 /) is suddenly applied to the 
amplifier of Prob. 8.1. 

(a) Write the equation for the output voltage v . 

(b) Sketch the envelope of the output signal v from t = to t = 100 /«ec. 

(c) Determine the rise time of the envelope. 

id) Discuss the relationship between the envelope rise time and the circuit Q. 

PROB. 8.3. Prove that the time response of the transistor amplifier of Fig. 8.1fc 

is given by 

i KI(\ -e-'l 2RC )sma> t 

when the input current is i t = / sin co t beginning at t = 0. 

The steady-state sinusoidal response of the amplifier may be determined 
from the transfer function, Eq. 8.14. Replacing s by jco, 2£co„ by l/RC, 
and co n 2 by 1/LC, we have 

G(jco) = - & ^ (8.36) 

RC LC 

Dividing both numerator and denominator of Eq. 8.36 by jco and multi- 
plying both numerator and denominator by RC, we find that 



But 



W-'<* 1+JaR c + WJaL) ( ° 7) 

R = Qco L=-2- (8.38) 

co„C 



Then, recognizing that g m R = K, the reference gain, 



GOV, = ~K . * - (8-39) 

G(jco) = -K -i (8.40) 

\co CO I 

Figure 8.6 is a sketch of the voltage gain as a function of to. The half- 
power frequencies co H and co L occur when the magnitude of the denomin- 
ator of Eq. 8.40 is equal to (2)^. Therefore 

e (^_i^)L| G (^_^)Li (8.41) 

\w co H f\ I \co co L /l 



308 

Then 



<°H _ C°o _ m o _ <°L 

(O co H co L <o 



Electronic Engineering 



(8.42) 



Rearranging terms, we have 



°>o ™o Q>L <°H 

m h + <»L _ 0>o(<»H + <°l) 
co (o H <o L 



f o. 



707 K 




Therefore, 



Fig. 8.6. Variation of gain as a function of at. 



(o H co L = co 2 



°>o = (o H m L y A 

Rearranging the first term of Eq. 8.41, we find that 

co H 2 - co* _ i 
m m H 

From a combination of Eq. 8.46 and 8.47, 



m a) H 



= 1 



(8.43) 
(8.44) 



(8.45) 
(8.46) 



(8.47) 



(8.48) 



Then 



J H 



Oij. = 



(8.49) 



Small Signal Tuned Amplifiers 309 

The bandwidth B is defined as <o H — co L . Then 

B = ^ (8.50) 

Q 

Note that the bandwidth is in radians per second when co is in radians per 
second. 

PROB. 8.4. Calculate the bandwidth of the amplifier of Prob. 8.1. Express this 
bandwidth in KHz. Answer: 32.3 KHz. 

The gain bandwidth product of the single tuned vacuum tube amplifier 
may be of interest. 

KB = g m R^ (8.51) 



But 



R = Qa)oL= -Q- (8.38) 

ft)„C 



Then 

KB = — (8.52) 

C 

This is the same gain bandwidth product as that of the R-C coupled 
amplifier (Section 7.5), providing that the shunt capacitance is the same in 
each case. It may seem peculiar that the tuned amplifier has the same 
gain-bandwidth product as its R-C amplifier counterpart, when the 
envelope rise time is twice as great as the comparable R-C amplifier. The 
reason for this behavior will be understood after the amplitude modulation 
process has been studied (Chapter 14). 

8.2 TAPPED TUNED CIRCUITS 

The single-tuned circuit discussed in the preceding paragraphs does not 
provide any means of impedance transformation. The output impedance 
of a pentode tube amplifier is fairly well matched to the input impedance 
of a following tube amplifier; therefore, when used to couple pentode 
tubes, the single-tuned circuit described would provide nearly maximum 
available power gain. The power gain of a triode tube or transistor 
amplifier circuit could be materially increased, however, if an impedance 
matching feature were included in the coupling device. A tap on the coil 
in the tuned circuit provides this feature. An illustration is provided in the 
transistor amplifier of Fig. 8.7. In the equivalent circuit, Fig. 8.76, R is 
the output resistance of the driving transistor, i?! is the series resistance of 



310 



Electronic Engineering 




'©*<>§ Vl =r 




(b) 
Fig. 8.7. A tuned coupling device which will provide impedance transformation. 



the coil, and R t is the input resistance of the following transistor stage. 
There are n 2 turns between the tap and the r-f ground terminal of the 
coil. 

To obtain maximum power transfer, the impedance looking back into 
the tapped tuned circuit should be equal to R t . If there were no power loss 
in the tuned circuit, maximum power transfer would be obtained also when 
the impedance looking into the collector side of the tuned circuit is R . 
Although there are actually no lossless tuned circuits available, the tuned 
circuit may be made mathematically lossless by transforming the coil 
resistance R 1 to its equivalent parallel value R p by the process previously 
discussed and then adding this equivalent resistance in parallel with the 
output resistance R . This technique is illustrated in Fig. 8.8. The 
modified output resistance R' in the high Q case, using Eq. 8.8, is 



*'.= 



R o + Qo w oL 



(8.53) 



where w is the desired resonant frequency. The value of the inductance L 
is not known at the moment, but will be determined later. Then R' can be 
calculated. 



311 



Small Signal Tuned Amplifiers 

The driving point impedance of the tapped coil and its load R { will now 
be found. Writing loop equations for the tapped coil and its load, we have 



- V ± = jcoLh - jco(L 2 + M)I 2 
= -ML, + M)h + (Ri +j<»L 2 )I 2 



(8.54) 



where M is the mutual inductance between L 2 and L x . Solving for I lt we 
find that 

-(R t + jcoL 2 )V 1 



1 jcoIXRi + jcoL 2 ) + a>\L 2 + Mf 
The driving point impedance of the tapped coil is 

-V 1 jwLjRj + jo)L 2 ) + ft> 2 (L 2 + Mf 



Zi = 



h 



R t + j(oL 2 



Z x = jcoL + 



co\L 2 + Mf 
Rt + jcoL 2 



(8.55) 



(8.56) 



(8.57) 



The second term on the right side of Eq. 8.57 is the impedance coupled into 
the coil because of the load R t . Since this term is complex, the effective 
reactance of the coil, and consequently the resonant frequency of the tuned 
circuit, will change with R f . This effect is highly undesirable because i?,- 
changes with bias, temperature, etc. It can be seen from Eq. 8.57 that the 
coupled impedance will be almost purely resistive if coL 2 is small in com- 
parison with R t . Actually, R { will always be large in comparison with a>L 2 
in the high Q case, since the circuit Q is the ratio of the effective shunt 
resistance to the shunt inductive reactance. Then the driving point 




(a) 




(b) 



Fig. 8.8. (a) The equivalent parallel resistance R v , which represents the power loss in 
coil L is combined with the transistor output resistance R to produce the modified 
output resistance R' in b. 



312 

impedance becomes, 



Z 1 ~ja>L + 



ay\U + Mf 



Electronic Engineering 



(8.58) 



In the usual coil configuration, the length of the winding space is short 
in comparison with the diameter of the coil. Then, the coefficient of 
coupling approaches unity, and the inductance of a coil is proportional to 
the number of turns squared. Then 



® 



(8.59) 



where « 2 is the number of turns in that part of the coil which is common to 
both input and output and n is the total number of turns in the coil. Also 



The mutual inductance is 



=■«(# 



M = k^/L^ = k 



Ln x ni 



Since n 1 = n — n 2 , 



Eq. 8.58 then becomes 



Z t ~ jcoL + 



Zj = jcoL + 



"-"(?-£) 
(^H?-^' 



R< 



2 L 2 r(i-fc)(^y+fc- 2 T 



R, 



(8.60) 



(8.61) 



(8.62) 



(8.63) 



This driving point impedance Z is the effective series impedance of the 
tapped coil and its load R { as shown in Fig. 8.9a. The right-hand term of 
Eq. 8.63 represents the equivalent series resistance R' s in the coil L resulting 
from the load R { . To obtain maximum power transfer into the load, this 
series resistance must appear as a parallel resistance equal to R' as shown 
in Fig. 8.96. Using Eq. 8.6 again to transform the effective series resistance 
into its equivalent parallel value, we see that 



R\ = 



(joLf 



(wLf 



co 2 L 2 



(1-fc) 



f- , T+* 5a T 
\n/ n- 



(8.64) 



R< 



Small Signal Tuned Amplifiers 

h 



313 



© > R 'o^F 



Ir; = 



'L'[(l-*)ffi 2 +*g) 2 




(a) 



(b) 



Fig. 8.9. The equivalent series resistance of (a) is transformed into the equivalent 
parallel resistance R' in (6). 



or 






(8.65) 
(8.66) 



Solving for — , we see that 
n 



«2 

n 



2(1 - Id L4(l - fe) ; 



r * WR'rr (8 . 67) 
L 4 (i - k? + i - k J 



The signs in Eq. 8.67 must be chosen so that n 2 /n is real and positive. Then 

».r *■ + w«'/]'^ ^_ (8 . 68) 

n L4(l - fc) 2 1 - k J 2(1 - fc) 

Remember that Eq. 8.68 is valid only for coils with fairly good magnetic 
coupling. 

A test coil could be wound on the desired coil form and the value of 
k determined for that particular configuration. For tightly coupled r-f 
coils, typical values of k range from about 0.8 to 0.9. It can be seen from 
Eq. 8.66 that when k approaches unity 






(8.69) 



As previously mentioned, the value of R' cannot be calculated until the 
coil inductance L is known, or vice versa. Two equations which involve 
both R' and L are Eq. 8.9 and Eq. 8.53. Since the effective shunt loading 
resistance on the tuned circuit is R'Jl in the maximum power transfer case 



314 Electronic Engineering 

(Fig. 8.9b), Eq. 8.9 can be written 

Y = Qco L (8.70) 

where Q is the circuit Q = cojB. Then 

R' = 2Qw L ' (8.71) 

Substituting this value of R' into Eq. 8.53, we have 

2Qco L = R " Qo0ioL (8.72) 

R + Q o co L 

2 q = M« (8 73) 

R» + Qo^„L 

2QR + 2QQ co L = R Q (8.74) 

l = m^m = &/j_ _ n (8 . 75) 

Observe that the inductance L determined by Eq. 8.75 will provide the 
desired bandwidth when maximum power transfer is achieved. Also 
notice that maximum power transfer cannot be achieved unless the coil Q 
is more than twice as great as the circuit Q. Since the circuit Q is usually 
determined by the bandwidth requirement, the minimum coil Q which 
will provide maximum power transfer can be quickly calculated once the 
resonant frequency and bandwidth are set. -Of course, the coupling 
efficiency and hence the total stage gain increases as the coil Q is increased 
above this minimum value. In the development it has been assumed that 
the loss in the tuning capacitor is negligible and the circuit Q is high (ten 
or more). 

The number of turns of wire for a cylindrical coil configuration can be 
determined from the empirical formula 

2 2 

L = — — — x 10- 6 (8.76) 

9r + 10/ v ' 

where r is the mean radius of the coil in inches, / is the length of the coil 
in inches, and L is the inductance in henries. After the coil form with its 
radius r and winding space / has been selected, the total number of turns 
n can be calculated from Eq. 8.76. Then the number of turns to the tap n 2 
can be calculated from Eq. 8.68. 

PROB. 8.5. A transistor with R„ = 10 KQ is to be coupled to a transistor with 
Ri =1KQ. The resonant frequency is 5 MHz and the desired bandwidth 



Small Signal Tuned Amplifiers 315 

is 200 KHz. Using a tapped tuned circuit to provide maximum power transfer, 
determine the total number of turns and the location of the tap. The coil is to 
have a Q„ = 100 and it is to be wound on a 0.2 inch diameter form and fill a 
0.25 inch winding length. Assume k =0.8. Answer: n = 32.8, n 2 = 16. 
PROB. 8.6. Assuming that k approaches unity, calculate n and n 2 for the coil of 
Prob. 8.5. Compare this coil with the one designed for k = 0.8. 
PROB. 8.7. A long, small diameter coil has a small value of k. Assuming k to 
be approximately zero so that L 2 = (nJn)L, derive an expression for njn in 
terms of R t and R'„. Compare the results with the tightly coupled case. 

The foregoing problems lead to the conclusion that sufficient accuracy 
may be obtained for most applications if Eq. 8.69, which was obtained for 




Fig. 8.10. An alternate impedance matching system. 

the case k ~ 1, is used for all degrees of coupling in the tapped coil. In 
this case, the tap location can be determined directly from a knowledge of 
the load resistance R t and the modified output resistance R' . 

The impedance transformation could have been accomplished by "tap- 
ping the capacitor" instead of the coil as shown in Fig. 8.10. The relation- 
ship between the capacitive reactances in this case would be the same as 
the relationship between the inductive reactances of the tapped coil where 
k = 0, (i.e., the reactance is proportional to the number of turns). 
Provision is always made to vary at least one of the tuning elements to 
tune the circuit. The nontapped element is the only single element which 
can be conveniently, varied without changing the impedance ratio. 

PROB. 8.8. The circuit of Fig. 8.10 is to be used to couple the transistors of 
Prob. 8.5. Calculate the value of each of the tuning capacitors. Answer: C x = 
5.75 pf, Q = 7.1 pf. 

8.3 INDUCTIVELY COUPLED SINGLE-TUNED CIRCUITS 

Impedance transformation may also be accomplished by inductively 
coupling the tuned circuit to either the source or the load. This method 



316 



Electronic Engineering 




(b) 

Fig. 8.11. (a) Transistor circuit and (6) equivalent circuit which accomplish impedance 
transformation by inductive coupling. 

is illustrated in the transistor circuit of Fig. 8. 1 1 . In reference to the 
equivalent circuit, Fig. 8.116, voltage equations can be written for the 
coupling circuit which is within the dashed enclosure. In this case, R { will 
be included as part of Z 22 . 



V 1 — l\Z-w + I2.Z12 

U = I\£~2\ -r / 2 Z 2 2 

where Z u =ja)L 1 + /? x 

Z 22 =jmL 2 + R 2 + R ( 

Z 12 = Z 21 = jcoM 
Solving for I u we have 

K 1 Z 2 2 



(8.77) 



A = 



■^11^22 ^12 



(8.78) 



Small Signal Tuned Amplifiers 

The driving point impedance of the primary is 



7 _ V l _ Z ' 
•»1 



I Zoo Zi 



= z„ - 



Substituting the values of Z u , Z 12 , and Z a2 above, 

(coM) 2 



Z ln = R t + }<»L\ + 



317 



(8.79) 



(8.80) 



R 2 + K, + J°>L 2 

The last term in Eq. 8.80 represents the impedance coupled into the 
primary. To achieve good efficiency and minimize the reactance coupled 
into the primary, R ( should be large in comparison with both R 2 and wL 2 . 
Then 

(wM) 2 



R t + jcoLi + 



(8.81) 



To provide good coupling efficiency, the coupled resistance should be 
large in comparison with the ohmic resistance of the primary -R x . If /?i is 
not negligibly small, its parallel equivalent can be lumped with R„ to 
provide R' as was done for the tapped tuned circuit. Then 

ijcoL 1 + y — 



Zm! 



R< 



(8.82) 



The simplified equivalent circuit of Fig. 8.12 can then be drawn. The 
parallel resistance of the tuned circuit at resonance must be equal to R' to 
provide maximum power transfer. Using Eq. 8.6 



R„ — R „ 



[(coMf/Rt] 



- ®- 



(8.83) 



Note the similarity between the transformation ratio of the tapped tuned 
circuit and the inductively coupled circuit. The latter has the advantage in 
most transistor circuits because the bias circuit may be isolated from the 
signal circuit, as illustrated in Fig. 8.11a. Higher current gain may 
therefore be achieved because the bias resistors do not absorb signal 
power. 




Fig. 8.12. A simplified equivalent circuit. 



31" Electronic Engineering 

The value of M = k{L^L^ may be substituted into Eq. 8.83. Then 

(8.84) 



K'» = -r*-K, 



Practical values for L 2 may be determined from the requirement that 
coLz be small in comparison with R ( (for example, one tenth) as previously 
mentioned. Also the value of L x will be determined by the bandwidth 
requirement of the amplifier. As for the tapped tuned circuit, the total 
shunt resistive load on the tuned circuit at the resonant frequency is R'J2. 




Fig. 8.13. An inductively coupled tapped circuit. 



Then 



Q = 



2co L 1 B 



(8.85) 



where B is the desired bandwidth in radians per second. This is the very 
same problem which was solved for the tapped tuned circuit. In that 
circuit, the inductance in the collector circuit was found to be 

After determining the proper values of L 2 and L u we may determine the 
proper coupling coefficient k by the use of Eq. 8.84. A preferable solution 
would be to determine experimentally the maximum realizable value of 
coupling coefficient k for the particular coupling coil under consideration. 
Then the required value of secondary inductance L 2 could be determined 
from Eq. 8.84. This procedure would yield the minimum value of Z, 2 and 
hence the minimum reactance coupled into the tuned circuit. 

It may be seen from Eq. 8.75 that L x must be very small if the required 
bandwidth is narrow and the desired circuit Q approaches the Q of the 
coil. Then the required tuning capacitance will be very large. The in- 
ductance L x may be increased and the tuning capacitance decreased if a 



Small Signal Tuned Amplifiers 319 

tapped tuned circuit is inductively coupled to the load as shown in Fig. 
8.13. 

PROB. 8.9. A transistor which has R = 20 Kfi is to be coupled to a transistor 
which has R ( = 1 Kfl. The center frequency is 1 MHz and the desired band- 
width is 20 KHz. Using an inductively coupled single tuned circuit (without 
tapping), calculate suitable values for L lt L 2 , and C. The Q„ of the coil is 150 
Assume k = 0.8. 

PROB. 8.10. Calculate suitable values for L l3 L 2 , and C when the collector of 
the transistor of Prob. 8.9 is connected to a center tap on the primary coil. (Q„ = 
150.) Assume that k remains 0.8. Answer: L x = 42.5 nhen, L 2 = 2.5 phen, 
C = 597 pf. 

The inductively coupled and tapped circuits could be utilized in vacuum 
tube circuits as well as transistor circuits. In this case the grid circuit is the 
higher impedance. Therefore, the untuned winding is normally in the 
plate circuit. Since the tuned winding may be used as a d-c grid return, R t 
is essentially infinite. Then the ohmic resistance of the coil cannot be 
neglected. Instead, it provides the load resistance of the coupling circuit. 

The plate resistance for a pentode tube is so large that an impedance 
match cannot usually be obtained with practical values of coupling. 
Satisfactory power gain may be obtained, however, with realizable 
coupling values. 

8.4 DOUBLE-TUNED CIRCUITS 

Two tuned circuits may be used as a double-tuned circuit to couple two 
amplifying devices. The tuned circuits may be coupled by either mutual 
inductance or mutual admittance, as shown in Fig. 8.14. The mutual 
admittance may be provided by either a capacitor or an inductor as shown. 
In any case, the degree of coupling or coupling coefficient is usually small 
in comparison with the values used in the single-tuned circuits previously 
discussed. The gain-bandwidth product may be higher in the double- 
tuned circuit and the designer has greater freedom in the frequency 
response characteristics as compared with the single-tuned circuit. These 
features will be discussed in the following paragraphs. 

To simplify writing of the equations, let 

G + sC + — =Y (8.86) 

sL 

and for the inductance coupled circuit, 



320 



Electronic Engineering 




r© 



'© 



Y Y 

(a) Mutual inductive coupling 






(b) Capacitive coupling 



Jbm 



c^= v 1 



>G V 2 



r y 

(c) Inductance coupling 
Fig. 8.14. Typical double-tuned circuit coupling techniques. 



Then the two nodal equations for the circuit are 
-I=(Y+ YJV,- Y m V i 
0=-Y m V 1 + (Y+ YJV 2 
These equations are solved simultaneously for V 2 . Thus 

-yj _ -yj 

(Y+Y m f-Y m 2 Y(Y+2YJ 



V* = 



(8.88a) 
(8.88b) 

(8.89) 



Small Signal Tuned Amplifiers 321 

When the values of Y and Y m are substituted into Eq. 8.89, V a is 

I_ 

V 2 = - sLm r (8.90) 

G + sC + — )[G + sC + \ + -7-) 

sL sL m / 



( G+sC+ i)( 

If we let L\L m = k, Eq. 8.90 becomes 



J 



or, 



vT 

y __ JX -'m to g]\ 

y _ *■■" *- (O Q2) 

\ C LC!\ C LC I 

When the common terms have been cancelled, Eq. 8.92 can be written as 

Ji. _ ^2. ! (8.93) 

-/ C (s - pj)(s - p 2 )(s - p a )(s - p 4 ) 

where co n 2 = 1/LC 

*■—-%*>&-$$—>*'*• <8Ma) 

G , ./l+2fc G 2 \* , _ 

= -ai±M(l + 2fc) M (8-95) 

The pole-zero plot of the transfer function F 2 // is shown in Fig. 8.15a. 
As noted in Chapter 7, the magnitude of the transfer function K 2 /Jis equal 
to the distance from a given w to the zero, divided by the product of the 
distances from the given to to each of the poles. For a high g-tuned 
circuit, interesting and useful values of co occur only in the vicinity of the 
poles p 1 and p 3 near the +jto axis. In this Vicinity, the distances to the 
zero and to the poles p 2 and p i near the — jco axis are large and almost 
constant for small variations of co. Therefore, when co is restricted to the 
region shown in Fig. 8.156, the transfer function can be written for steady 
state sinusoidal signals as 

VJI = — (8.96) 

(jco - Pi)0'« - p a ) 



322 



Electronic Engineering 



P3X 

PiXH 



P2X 



(a) 



3 






K P3 



p\ 



v, 
V 



c 



<S\- 



(b) 



Fig. 8.15. A plot of the poles and zeros for the function VJI of Eq. 8.93. 



where A: is a constant to be evaluated. The frequency is approximately 
co in this region. When this value of w is substituted into Eq. 8.93, and 
it is recognized that w n ~ w and (jco - p 2 ) ~ (jw - Pi ) ~ 2w , Eq. 
8.93 becomes 



1 



ku>„ 



1 



AC (jw - pjijco - p 3 ) 



(8.97) 



The expression for the pole p 3 = —o l +jw \ / l + 2k 2 can be simplified 
by the use of the binomial theorem. Since in practical double-tuned 
circuits, fc« 1, o nly the first two terms of the binomial expansion are 
significant and sfl +2k 2 ~\ + k. Then, p 3 ~ -a x + jio {\ + k), and 
the two poles are separated by ka> (Fig. 8.156). Also, since the area of 
the triangle co Pl p 3 (Fig. 8.156) is \ sin <f> \jw - Pl \ \jw - p 3 \ and this area is 
also \a-Ji(x> , 



li'w - Pil \}<o - p 3 \ = ^-^ 
sin <f> 



(8.98) 



Substituting Eq. 8.98 into Eq. 8.97, the magnitude of the transfer function is 



(8.99) 



sin <f> sin <f> 
~ 4ff x C ~ 2G 

Equation 8.99 shows that maximum output voltage K 2 is obtained when 
<f> = 77/2 and sin <f> — 1. Then (since / = g m V gk ) the reference voltage 
gain of a tube amplifier is 



K., = 



(8.100) 



Small Signal Tuned Amplifiers 



323 



The locus of points which gives <f> = tt/2 is a semicircle with kw as a 
diameter (Fig. 8.16). Observe that there may be two frequencies (values 
of joy) which yield maximum voltage output and hence maximum power 
transfer. These are the frequencies at which the maximum voltage locus 
crosses the jco axis. There will be a single max- 
imum power frequency when the maximum J<" 
voltage locus is tangent to they'co axis. Then 



= a, = 



G_ 

2C 



(8.101) 



X«5-.. 



where k c is the value of k which produces this 
single maximum power frequency and is called 
the critical coupling. Then 



fc, = 



G 

<w„C 



1_ 

Q 



(8.102) 



'/ 
X— 

L 



oi 






Maximum 

voltage 

locus 



Fig. 8.16. Details of the 
pole-zero plot of Fig. 
8.15 near P r and P,. 



When the coupling is less than critical, there is 
no frequency at which theoretical maximum 
power transfer occurs. When the coupling is 
greater than critical, there are two frequencies at 
which maximum power transfer occurs, as 

previously discussed. The coefficient of coupling is often given in terms 
of the critical coupling k c . For example k = bk c . A sketch of the fre- 
quency response for several values of b is given in Fig. 8.17. Note that 
one pole remains fixed at co and the other pole moves. The separation 
between poles is bk e <o and the center of the pass band is 



o c = o> + 



bk. 



= «\,[l + 



bk r . 



(8.103) 



2.5^. 



'^<c- 



2.01 \ 

b iL \rt~ 

0.5k Y-htr- 
/// 



'^ 




^V 2 



Fig. 8.17. A sketch of the frequency response for several values of relative coupling b. 



jta 



u 



Jk 



i 



i A 



i 



11/ 
i / 

3\c 



I 

L 



*c<% 



\ 
\ 



<0 H 



324 Electronic Engineering 

The bandwidth may be determined as a function of b with the aid of 
Fig. 8.18. Since the output voltage is proportional to sin <f>, as shown in 

Eq. 8.99, the half-power frequencies co H and 
oi L will occur when sin <f> = 0.707 or <f> = tt/4 
radians (Fig. 8.18). In this figure, the base 
of the trapezoid is cd h — <o L = bandwidth B. 
To determine the bandwidth, we note that 
from Eq. 8.101: 

o x = ±f (8.101) 

From Eqs. 8.98 and 8.101 the product of the 
two sides C and D of the triangles shown in 
Fig. 8.18 is 

CD = ^2- 2 (8.104) 

Observe that the area of the triangle with sides 
B, C, and D (Fig. 8.18) is, using Eq. 8.101, 

k e co B 



B 



-J 



A = ia x B = 



(8.105) 



Fig. 8.18. The configuration 
used to determine the band- 
width of a double tuned 
circuit. 



where B is the bandwidth m H — <o L . Also the 
area of this triangle may be determined from 
the following relationship: 

A = \CD sin a 



Equating Eqs. 8.106 and 8.105, the bandwidth is 

2CD sin a 



B = 



^o 



(8.106) 



(8.107) 



Substituting the value of CD from Eq. 8.104, 



B = (2) A b sin afc c «> = (2) A b sin <x =2 
The angle a may be determined from Fig. 8.18 



(8.108) 



a = 225° - 2 sin -1 



}A-\ \'A 



K^-pmi 



(8.109) 



Equation 8.109 is valid only for values of b between 1 and approximately 
2.5 because maximum power transfer was assumed in the derivation, and 



Small Signal Tuned Amplifiers 



325 



the pass band splits in two at b ~ 2.5 (Fig. 8.17). Since the angle « is 
rather tedious to evaluate, the values for b = \, 1.5, and 2.0 are given in 
Table 8.1. Interpolation may be used to obtain sin a for other commonly 
used values of b. 





TABLE 


8.1 


b 


a 


sin a 


1.0 


90° 


1.0 


1.5 


104° 


0.96 


2.0 


113° 


0.92 



Note that the term (2) lA b sin a represents the improvement in the gain- 
bandwidth product of the double-tuned amplifier over the single tuned. 
It should be observed that the Q used in Eq. 8.108 is the loaded Q of 
each tuned circuit, not considering the coupled resistance. This is in 
contrast to the single-tuned circuit where the Q must include all loading, 
including the coupled resistance. 




nVQ 



*o 




(a) (b) 

Fig. 8.19. A double-tuned capacitively coupled transistor amplifier. 

The double-tuned circuit which uses the coupling inductance L m = Ljk 
yields readily to analysis, but it is not the most convenient type of coupling. 
A capacitor C m which has mutual admittance jwC m is smaller and less 
expensive then the inductor and has the additional advantage of blocking 
direct current. The circuit diagram of a capacitively coupled double- 
tuned transistor amplifier is given in Fig. 8.19a. Because of the similarity 
of capacitive coupling to the inductive coupling previously discussed, we 
can see that the relationship between the coupling capacitance and the 
tuning capacitance is 

Cm = kC = bCIQ (8.110) 



326 



Electronic Engineering 



The analysis of the capacitively coupled circuit will show that the 
fixed pole is at a> , but the movable pole is below w and moves downward. 
The band center is, therefore, below a> . However, the bandwidth is 
usually so small in comparison with w that the band center can be 
considered to be at w in most applications. The movable pole does not 
move parallel to they'w axis when capacitive coupling is used. Therefore, 
the analysis of this circuit is somewhat more complex than the analysis 
of the inductor coupled circuit. However, in the high Q case, the pole 
motion is approximately parallel to they'w axis and the preceding analysis 
of the inductor coupled circuit is applicable to the capacitively coupled 
circuit. 




Fig. 8.20. A double-tuned transistor circuit. 

The equivalent circuit of Fig. 8.19ft shows that the input resistance R f 
must be transformed to R , not R' , in order to provide equal Q in both 
primary and secondary circuits. Therefore 

(8.111) 



»« \rJ 



Mutual inductance may also be used to couple the coils of a double- 
tuned circuit (Fig. 8.20). The mutual impedance is then jcoM . The value 
of critical coupling coefficient k c is 1/Q and, in general, k = bk c = bjQ 
or M = bL/Q. 

The mutually coupled circuit is most easily analyzed if both primary 
and secondary circuits are converted to equivalent series circuits and loop 
equations are used. The complex poles are then symmetrically located 
on either side of co and they both move away from w as k is increased. 
The poles do not move parallel to they'eo axis. However, the analysis of 
the inductively coupled circuit is adequately applicable to the mutually 
coupled circuit. 

When the coil Q is high in comparison with the circuit Q, the major 
portion of source energy is transferred to the load. This follows from the 



Small Signal Tuned Amplifiers 327 

fact that the effective series resistance caused by source and load resistances 
is large in comparison with the series ohmic resistance. When an imped- 
ance match is obtained, the power delivered to a specific load impedance is 
essentially independent of the type of coupling circuit, provided the coup- 
ling efficiency is high. The various circuits described in this chapter will 
each produce about the same reference gain under matched conditions. 
The double-tuned circuit will have the largest gain-bandwidth product 
because of the bandwidth ratio expressed in Eq. 8.108. In addition, the 
frequency response characteristics of the double-tuned circuit have steeper 




Fig. 8.21. Typical time domain response of a double-tuned amplifier driven by a 
rectangular pulse of frequency o> . 

sides and a flatter top than the single-tuned circuit. Also the designer has 
some control on the shape of the response curve by his choice of b. The 
best choice would depend on the application. Values for b of the order of 
1.0 to 1.7 look promising for cases where uniform frequency response is 
desired over most of the pass band. 

The development of the transient or time domain response of a double 
tuned amplifier is beyond the scope of this book. However, Martin 2 has 
shown that the double-tuned amplifier (or actually its staggered pair equiv- 
alent) has an overshoot in the envelope response which is comparable with 
the overshoot of a transformer coupled amplifier. Thus, the improved 
gain-bandwidth product or gain-rise time ratio of the double-tuned ampli- 
fier is obtained at the expense of overshoot in the envelope response. This 
overshoot is illustrated in Fig. 8.21 where it is assumed that the excitation 
is a pulse-modulated sinusoid. Martin showed that the critically or 
transitionally coupled amplifier has a 4.3 % overshoot. It has also been 
shown that the overshoot increases as the degree of coupling increases. 

2 Thomas L. Martin, Jr., Electronic Circuits, Sec. 6.6, Prentice-Hall, Englewood 
Cliffs, N.J., 1955. 



328 Electronic Engineering 

Therefore, the degree of coupling b which provides the optimum frequency 
response characteristic does not, in general, provide optimum time domain 
response. 
The procedure for designing a double-tuned circuit might be as follows : 

1. Choose b for the best response characteristics. 

2. With a> and the desired bandwidth known, calculate the circuit Q from Eq. 
8.108. 

3. Calculate the coefficient of coupling 

k = bk c = I- 
c Q 

4. Determine the primary inductance from the relationship Q = R'JoigL, where 
the output resistance R' includes the effective parallel resistance of the primary 
coil. However, as for the tapped tuned circuit and the inductively coupled 
tuned circuit, the effective parallel resistance of the primary coil cannot be 
determined until the primary inductance is known. Therefore, the relation- 
ship stated previously (Q = R' jui o L) must be used in addition to the relation- 
ship R' = R v R j(R p + R ) to calculate the primary inductance. Using these 
two relationships, the value of /?'„ becomes 



R P R 











•" \ 


R P + R 


But R„ = 


G„* 


oL. 


Then 


Q<°oL 


QoOioLRo 
Q ca L + R 


Simplifying, 




















Q(Q„<»oL + R ) = Q„R„ 



(8.112) 



(8.113) 



L = 



(8.114) 
QQ o>„L = R (Q - Q) (8.115) 



The secondary inductance would, of course, be the same as the primary 
inductance if this suggested design technique is used. 

5. Calculate the value of capacitance required to resonate with the inductance 
calculated in step 4. 

6. Determine the necessary tap points on the coil to provide proper loading of 
the tuned circuits. 

An example of an inductively coupled double-tuned transistor amplifier 
will be given to illustrate this outlined procedure. 

Example 8.1. A double-tuned circuit is to be used to couple a transistor which 
has R = 40 KH to a transistor which has Ri = 1 KO. The center frequency is 
500 KHz and the desired bandwidth is 12 KHz. The foregoing suggested pro- 
cedure will be used in designing the coupling circuit. 



Small Signal Tuned Amplifiers 329 

1. The coupling coefficient 1.5 k c will be considered optimum (b = 1.5). 

500 KHz 

2. B = 12 KHz = 1.41 x 1.5 x 0.98 

Q 
2.06 x 500 KHz 

3 - g - 12 KHz ~ 86 

b 1.5 

* = G=T6= - 0175 



4. Assuming the coil Q = 150. 



40 x 10 3 



^ - *■ - 3.14x10* '^ " TTS) " « A* 



SC= (3.14 x 10«W3 x 10-0) " "10 pf 
6. The tap on the secondary coil is 

l V* 



\86 150/ 



"*^ n (jd) =0158w 



The inductance could be increased and the tuning capacitance reduced from the 
values just calculated by tapping the primary as well as the secondary. The 
problem would then proceed as before but with a transformed value of R . 

PROB. 8.11. A transistor having R = 20 Kii is to be coupled to a transistor 
having R { = 1 KQ. The desire! bandwidth is 20 KHz and f = 460 KHz. 
Design a double-tuned circuit using b = 1.2 and C = 100. 
PROB. 8.12. Use pentode tubes with r„ = 1 megohm for the amplifier of Prob. 
8.11. For good frequency stability assume C = 100 pf. Use loading resistors 
to reduce the load resistance of both primary and secondary. Answer: Q — 38.3, 
L = 1.2 mh, R = 215 Kto. 

PROB. 8.13. The circuit pf Fig. 8.19 is used with f = 460 KHz and band- 
width = 20 KHz. Using e = 1.5 and G' = G' t = 50/<mhos determine Q x = 
Q 2 and C m for the circuit. The Q of each coil is 100. 

8.5 ' NEUTRALIZATION ?6<Z*SA 

It was learned in Chapter 7 that amplifiers such as triode tubes and 
transistors which have a ppreciable capacitance between their input an d 
output circuits may be unstable at high frequencies. This instability 
occurs when the input an d outp ut circuits are inductive. Since a parallel - 
tuned circuit is inductive at trequencies below resonance, tuned amplifiers 
which use triodes or transisto rs are especially susc eptible to oscillation. 

In addition to being unstable, amplifiers which have appreciable coupling 
between their input and output circuits are difficult to design, as previously 



330 



Electronic Engineering 



mentioned. Also, in the tuned amplifier, the reactance of the tuned 
circuit in the output affects the resonant frequency of a tuned circuit in the 
input and vice versa. Thus, tuning becomes very difficult. 

As was previously mentioned, a process known as neutralization may be 
used to eliminate the effects of coupling between the input and output cir- 
cuits of an amplifier. Neutralization is acco mp lished by providing a feed - 
back curr ent of opposite p olarity and equal magnitude to that feedback 
t hrough the amplifier. These two feedback currents cancel or neutralize 
each other, thus the name ne utralization! A neutralized tuned tnode 
amplifier is shown in hig. 8.22. When the coil in the tuned plate circuit is 
center tapped, the neutralizing capacitor C„ should have the same capaci- 




Flg. 8.22. A neutralized triode amplifier. 

tance as the grid to plate capacitance. When the coil is tapped at a point 
other than the center, the required neutralizing capacitance will be changed 
by the ratio of the signal voltages at the top and bottom of the tapped 
tuned circuit. The neutralizing voltage could have been obtained from the 
secondary of an inductively coupled circuit utilizing the phase reversing 
capabilities of the circuit. 

The neutralization of a transistor is not as straightforward as that of the 
vacuum tube because of the base spreading resistance in the transistor. 

To neutralize the transistor, a series R-C combination may be needed 
in the neutralizing circuit as shown in the hybrid-rr circuit of Fig. 8.23. In 
this circuit r c has been omitted as it frequently may be. It may be seen 
that neutralization will occur if /„ = I b . Then there will be no current flow 
and hence no voltage V\ in the input circuit resulting from an output 
voltage V 2 . Assuming the voltage to be sinusoidal, the current I b will first 
be determined. Considering the circuit to be neutralized so that V 1 = 0, 
we find that 



/„ = 



(8.117) 



Small Signal Tuned Amplifiers 

but 

'b'e = 

where 



*•/(*/.+ !) + *» +;'">£! 



331 



(8.118) 



1 a l 

g e = - and g b = - 



= — vwwv- 



Jb 



-A/WWV 



Vl (h fe +l)r e % i V b . e © 



.^2 
a 



V 2 



.rd 



Fig. 8.23. A neutralizing circuit for the common emitter configuration. 

Fig. 8.23 shows that the feedback current is 

V. 



h = 



1 



+ 



g.l(h f . + 1) + -g b + J°>Ci J mC o 

j<oC t (-^—+-g t +J<oC 1 )v t 

\h fe + 1 l__ 



(8.119) 



jcoC c + —^— + g b +j<oC 1 

hfe + 1 

Substituting Eq. 8.119 into Eq. 8.118, we have 



'h'f. 



Ve jcoC c + gj(h fe + 1) + g b + jcuCj 
and, using Eq. 8.117, we have 



(8.120) 



/„ = 



ja>C c V 2 



r b [jcoC c + jmd + g b + geKKe + 1)1 



jcoC c V 2 



jcor b {C c + C x ) + 1 + r bg J(h u + 1) 



(8.121) 



(8.123) 



332 Electronic Engineering 

Solving for /„ (see Fig. 8.23), we see that 

I n Ysb. (8>12 2) 

R n + (llJwC n ) 

where a is the ratio of the output voltage V 2 to the voltage applied to the 
neutralizing circuit. Equating /„ and /„, we have 

VJa jcoC c V 2 

R n + (llJcoC n ) jcor b (C c + CO + 1 + r b gj(h u + 1) 
Then 

^+-1— ir rt (gL±^ + i + ^*/.+D i (8 . 124) 

jcoC n at C c ja)C c J 

Equating reals and imaginaries, we have 

Since co a = \\r e C x , the neutralizing resistor R n may be expressed in terms 
of r e and co a instead of Q. 

C n = -. ^ (8.126) 



(1 + -S&-) 

\ K + 1/ 



Transistors such as the drift-field type which are designed especially 
for high-frequency applications often have a low value of ohmic base 
resistance and also a small collector-base junction capacitance. This 
combination may permit satisfactory neutralization with R n = 0. The 
neutralizing capacitance C„ is then equal to aC c where C c is approximately 
equal to the C ob listed by transistor manufacturers. 

PROB. 8.14. A 2N384 transistor is used as an amplifier at 10 MHz. 
Inductive coupling is used in the output with a ratio of two primary turns to each 
secondary turn. The coefficient of coupling approaches unity. Determine the 
neutralizing components for the amplifier, using the secondary voltage as the 
neutralizing voltage. Answer: R n = 1.75 K, C n = 2.5 pf. 
PROB. 8.15. Could a parallel R-C or R-L combination be used in the neutraliz- 
ing circuit of the amplifier of Prob. 8.14? If so, would it be as satisfactory as the 
series R-C combination ? Why ? 

8.6 THE RELATIONSHIP BETWEEN GAIN AND STABILITY 
IN A NONNEUTRALIZED TUNED AMPLIFIER 

Any amplifier will be stable if its power gain is sufficiently low. There- 
fore, neutralization will not be needed in a tuned amplifier if the amplifier 
gain is sufficiently restricted. The maximum usable value of gain for a 



Small Signal Tuned Amplifiers 



333 



nonneutralized amplifier may be approximately determined as a function 
of the feedback capacitance C c and the frequency of operation. The 
equivalent circuit of Fig. 8.24 will be used in this determination. This 
circuit represents either a tuned tube circuit or a tuned transistor circuit 
in which r b is negligible. The circuit components L, C, and G include the 
shunt device parameters, and the input circuit values are initially assumed 



'CD 



Vi 



\L zhC 



gmVd 



Q 



•G d±c 



Y Y 

Fig. 8.24. Equivalent circuit used for determination of maximum nonneutralized gain. 

to be equal to the output circuit values. Writing nodal equations for the 
circuit, 

I = (Y + sCM - sC c V 



~g m V< = -sC t V t + (Y + sC c )V 
Solving these equations simultaneously, the value of V„ is 

(sC c - g m )I 



V„ = 



Simplifying, 



(r+sC c ) 2 -sC c (sC c -sJ 
(sC c - gjl 



V„ = 



y» + 2sc.y+jc rf . 



(8.127) 
(8.128) 

(8.129) 
(8.130) 



Poles of V (and thus possible oscillation) will exist when the denominator 
of Eq. 8.130 is equal to zero. Therefore, the roots of the denominator 
will be found. Using the quadratic formula, the roots Y t and Y 2 are 



K* - ~ sC c ± vWj* - sC egm = -sC c ± JsC e (sC c - gj (8.131) 

In order for the amplifier to have appreciable power gain, the forward 
transconductance g m must be large in comparison with the passive mutual 
admittance sC c . Therefore, Eq. 8.131 may be simplified to 



and 



Y lt ~-sC e ±jy/sC egn 
-g m I 



(Y + sC e + j\/sCJJ(Y+ sC c - jy/sCaJ 



(8.132) 
(8.133) 



334 Electronic Engineering 

Substituting the value of Y = sC + G + 1/sL into Eq. 8.133, 



H 



-g m I 



sC + G-\ h sC c + 

sL 



J-JsC c g m \ ( 



sC + G-\ h sC c 

sL 



or 



-gfj 



s(C + C C ) + G + 



sL. 



-Jy/sC c g m j 

(8.134) 
(8.135) 



+ sC c g„ 



Let C + C c = C t . Then Eq. 8.135 becomes 



V n = 



[s 2 LC t + sLG + l] 2 + s 3 L 2 C cgn 



or 



K„ = 



-grJs 2 



ci 



s 2 + 



G ± ll 1 
C t LCj 



(8.136) 
(8.137) 



+ s 3 C c g„ 



Poles of V B occur when the denominator becomes zero, or 

s ^cgm 



and 



2 G 1 

s 2 + —s + 

C t LC t . 



C? 



s 3 C c g„ 



(8.138) 
(8.139) 



/ 1 G 2 V^ 

where co n = I — — - 1 

\LC t 4C*J 

This fourth-order equation may be reduced to a second-order equation 
if the circuit Q is high. Then the values of 5 which are of interest lie in 
the passband near a> . Since co is much larger than G/2C t , ja> may be 
substituted for either s or (s + G/2C t ) in the sum or product terms. 
Then 

(2/o>„) ^s + — - jco J = — (8.140) 

or 



, G . JWoC c g„ 



(8.141) 



Equation 8.141 involves the poles of V which lie near the +ja> axis 
near a> . The locus of these poles may be determined as the product 



Small Signal Tuned Amplifiers 

C c g m is varied. Rearranging Eq. 8.141, we have 



s + 



G ± (oj C c gJ ^e 
2C t 



l A „-j45° 



— ja> = 



335 



(8-14?) 



JO) 



The pole-zero plot of Eq. 8.142 is given in Fig. 8.25. Observe that a 
double pole occurs at s = — G/2C t +jco when C c = 0, but as C e is 
increased the poles move in opposite directions along a locus line which 
forms an angle of —45° with the real axis. 

As seen from either the root-locus plot or Eq. 
8.142, a pole will lie on the ja) axis (an undamped 
natural response will occur) when the real part of 
the lower pole vanishes. Then 

G = (co C cg J A cos 45° 
or 

2G 2 



(8.143) 



\ 



(8.144) 



U>o 



G 

2Q 



Fig. 8.25. The poles 
of V which lie near 
the +j<o axis. 



The value of transconductance determined by Eq. 
8.144 is the upper limiting value for a shunt load 
conductance G, coupling capacitor C c , and band 
center frequency w„. Any smaller value of g m will 
result in stable operation. However, Fig. 8.25 shows 
that the frequency response will be that of a nonsym- 
metrical double tuned circuit. The lower pole will 
appear as a higher- Q pole (nearer the/co axis) than 
the upper pole. In fact, as the lower frequency pole approaches the jm 
axis it will almost completely dominate the frequency characteristics and 
the amplifier will appear as a very narrow band amplifier. Therefore, if 
uniform, broadband amplification is desired, the conductance G should be 
adjusted so that the value of g m actually provided by the tube or transistor 
is much lower than the limiting value. 

The values of G seen from the input and output terminals of a transistor 
usually differ widely. Since both G's are equally important in determining 
the limiting value of g m , it is reasonable to replace G 2 in Eq. 8.144 with 
G'fi',,. Making this substitution, the limiting value of g m is 



1G\G' 

o„C c 



(8.145) 



PROB. 8.16. A given high-frequency transistor has C c = 10 pf. The total 
shunt resistances of the input and output circuits are 100 n and 1 Kfi, respec- 
tively. Assuming G G' ' t = G 2 , determine the limiting value of# m and the upper 
limit of collector Q-point current for stable operation at/„ = 10 MHz. Assume 



336 Electronic Engineering 

the transistor temperature to be 25°C. Answer: g m =3.18 x 10~ 2 , 7 cmax = 
0.8 ma. 

PROB. 8.17. A given transistor amplifier which has a quiescent collector 
current of 5 ma and a center frequency of 2.0 MHz is driven by a 200 H source 
at the resonant frequency. The transistor has a collector-base capacitance of 
5 pf. Assume T = 300°K. 

(a) What is the limiting value of output conductance G'„? 

(b) If, for stability and desirable frequency response characteristics, the 
output conductance G' is increased to three times the limiting value, what will 
be the approximate voltage gain of the amplifier ? Assume that r b ~ 0. 
PROB. 8.18. A single-tuned circuit is to be used, along with other necessary 
components, to couple two pentode tubes. Both pentodes are of the same 
type, having g m = 3000 /^mhos and r v = 1 megohm at the chosen Q point. 
Also these tubes have Cm = Com = 10 pf and C gP = 0.003 pf. The resonant 
frequency of the amplifier is 1 MHz. The tubes provide the total tuning 
capacitance. Calculate the voltage gain and the bandwidth of the stage if the 
coil Q = 100 and all grid resistors = 1 megohm. 

PROB. 8.19. A transistor having an output resistance of 20 Kft is inductively 
coupled to a transistor having 1 Kfl input resistance. The transformer primary is 
tuned to 1 MHz. The coefficient of coupling k = 0.7 and the primary coil 
Q = 150 at the resonant frequency of 1 MHz. Calculate the values of primary 
and secondary inductance required to provide maximum power transfer and 
25 KHz bandwidth. 

PROB. 8.20. A 1 MHz carrier is applied to the input of the amplifier of Prob. 
8.19 at t =0. Sketch the time response for the first few microseconds and 
determine the envelope rise time (10%-90%). 

PROB. 8.21. A double-tuned 460-KHz transistor i-f amplifier is to have a 
20-KHz bandwidth. The desired relative coupling b = 1.4. The coil Q = 120 
(both primary and secondary). The output resistance of the driving transistor is 
20 Kii, and the transformer secondary is tapped so Q 1 — Q 2 . Determine the 
coefficient of coupling k and the coil inductance L x = L 2 required to provide 
the desired bandwidth and the desired shape of the frequency-response curve. 



9 



Cascaded Amplifiers 



Most practical amplifiers require more gain than can be obtained from 
a single stage. Consequently, it is common practice to feed the output of 
one amplifier stage into the input of the next stage (Fig. 9.1). When 
amplifiers are connected in this fashion, they are called cascaded amplifiers. 
There are a few concepts, unique to cascaded amplifiers, which will be 
considered in this chapter. 

9.1 GAIN AND BANDWIDTH CONSIDERATIONS IN 
CASCADED AMPLIFIERS 

Most of the cascaded amplifier stages are used to obtain either a voltage 
gain or a current gain. However, in most cascaded amplifiers, it is 
ultimately the power gain that is important. When the proper level of 
signal has been obtained, a power amplifier stage (to be discussed in 
Chapter 10) is used to produce sufficient power to activate the required 
load device (loud speaker — servo motor — antenna, etc.). If a voltage 
gain is required, we can calculate the total gain by using the equation for 
voltage gain of one stage. Thus, from Fig. 9.1, the voltage gain for 
stage 1 is 

Gi = ;7 (9.D 

337 



338 



Electronic Engineering 



n 
























— >- 


-o 
-o 


Amplifier 

stage 

1 


o- 
o- 




>- 


-o 
-o 


Amplifier 

stage 

2 


o- 
o- 








t 

/l in 


out 


V 2 in 


out 


V 3 in 





















Fig. 9.1. Cascaded amplifier stages. 
In addition, the voltage gain for stage 2 is 

G 2 = - (9.2) 

V 2 

The gain for additional stages can be written in a similar manner. Then, 
the total amplifier voltage gain G A for n cascaded stages is 



Vt V 3 V 4 
— x — x — x • 

V 1 V 2 V 3 



or 



G A = G 1 xG 2 x G 3 X •■■G n 



(9.3) 
(9.4) 



Obviously, a similar derivation could have been achieved for current 
gains or for power gains. In either of these cases, the total amplifier gain 
is equal to the product of the individual stage gains as indicated by Eq. 
9.4. In general, the individual stage gains are functions of s and, con- 
sequently, the amplifier gain G A is also a function of s. For steady-state 
sinusoidal signals, s becomes jco and G u G 2 , etc., become magnitudes at 
given phase angles. Then, G A will be equal in magnitude to the products 
of all the magnitudes with a phase angle equal to the sum of the individual 
stage phase shifts. 

Most amplifiers are fundamentally band-pass amplifiers. Even the R-C 
coupled amplifiers are actually band-pass amplifiers since they do not 
pass frequencies all the way down to zero. (Amplifiers which do pass d-c 
signals and, therefore, can be considered as low-pass amplifiers will be 
considered in more detail in Section 9.5.) In these band-pass amplifiers, 
a reference gain K has been defined. Using this reference gain, an amplifier 
reference gain K A can be defined as 



±K A = (±K 1 ) x (±K 2 ) x (±K 3 ) x ■■ (±K n ) 



(9.5) 



The term ±K , will be positive if the total number of phase inversions is 
even and negative if the number of inversions is odd. Since ATis a magnitude 
only, this relationship does not involve s. 



Cascaded Amplifiers 339 

We have previously defined ft> x and <u 2 as the lower and upper cut-off 
frequencies, respectively. These are the frequencies at which the voltage 
or current gain of one stage has been reduced to 0.707 of its reference 
value. (Power gain is reduced to 0.50 of its reference value for resistive 
loads.) Now, if we have an amplifier with two identical stages of ampli- 
fication, the voltage gain at Wj will be reduced by a factor of 0.707 in each 
stage. Thus, the amplifier gain at a^ (and also co 2 ) will be 

0.707(-A„) x 0.707(-#„ 2 ) = 0.5K A (9.6) 

In fact, for n-identical cascaded stages of amplification, the gain at co 1 
and w 2 will be (0JQ7)"K A . 

To re-establish a meaningful amplifier bandwidth, let us define m L as 
the lower cutoff frequency of the cascaded amplifier and co H as the upper 
cutoff frequency of the cascaded amplifier. At these frequencies, the gain 
of the amplifier will be Q.1Q1K A . In order to arrive at some relationship 
between w x and w L and also between co 2 and m H , consider an R-C coupled 
amplifier containing n identical cascaded stages. The voltage gain per 
stage is given for the low frequencies by Eq. 7.19 as 

G=-K-^— (7.19) 

S + (Ox 

For sinusoidal steady state, s = jm. Then, Eq. 7.19 can be written as 

G = -K JW (9-7) 

JO) + «! 

or 

G=-K (9-8) 

a> 
The magnitude of this function can be written as 

™- K WT^ (9 ' 9) 

If there are n cascaded stages, the magnitude of the amplifier gain from 
Eq. 9.4 is 

IGJ - |G|" =*"( ■. , ' , 2 J (9-10) 

\[1 + (coi/co) J / 

Since K" = K A , we can write 



340 Electronic Engineering 

Now, if co is to be equal to co L , the term multiplying K A must be equal 
to 0.707 or 1/(2)*. Then H 

*"-['+ (MP 



or 



2 v. = i + ("lV (9 . 13) 

This equation is solved for co L 2 to yield 



- 2 i/» - 1 
or 



ft>i 



(9.14) 
(9.15) 



[2 1/n - 1] 
A similar solution of Eq. 7.55 yields 

co H = a) 2 [pl» - 1]« (9.16) 

PROB. 9.1. Derive Eq. 9.16 from Eq. 7.55. 

PROB. 9.2. Determine the value of co L /w 1 for n = 1, n = 2, « = 4, « = 8, and 
/I - 16. Make a plot of coj^ vs «. Answer: For n = 4, a> L la, l = 2.3. 
PROB. 9.3. Determine the value of m H /<o 2 for n = 1, n = 2, w = 4, « = 8, 
and n = 16. Make a plot of (o H loa 2 vs «. 

From these equations, co L will be greater than a, if n is greater than 
one and co H will be less than w 2 is n is greater than one. Thus, the band- 
width of the amplifier decreases as the number of cascaded stages increases. 
Or, if the amplifier bandwidth is to remain constant, the stage bandwidth 
must increase as the number of cascaded stages increases. This last 
statement leads to an interesting dilemma. If a high-gain very wide band 
amplifier is desired, stages must be cascaded to obtain the higher gain. 
However, as more stages are cascaded, the bandwidth of each stage must 
be increased. Unfortunately, as noted in Section 7.7, the gain-bandwidth 
product may be a constant. Under these conditions, the increased 
bandwidth results in reduced gain per stage. Thus, in order to compensate 
for the reduced gain per stage, more stages of greater bandwidth are 
required. This process can be carried so far that the total amplifier gain 
(for a given bandwidth) may actually decrease as additional cascaded 
stages are added. (This statement is only true when we are actually 
utilizing the maximum gain bandwidth possible from an amplifier stage. 
In most amplifiers, the bandwidth can be increased without decreasing 
the gain.) 



Cascaded Amplifiers ^4* 

From the foregoing analysis, a relationship between total reference 
amplifier gain and bandwidth is highly desirable. In a typical R-C coupled 
stage, «! « a> % . Thus, the bandwidth a> 2 - <% <** <o 2 . The gain-band- 
width product (or figure of merit for the stage) GB, can then be written as 

GB «s Kco 2 (9.17) 

or 

K = — (9.18) 

0) 2 

When an amplifier contains n identical cascaded stages, the reference 
amplifier gain K A is equal to 

K A = K n = ^ (9.19) 

<o 2 



However, from Eq. 9.16, we note that 



eo« = 



(9.20) 



[2 i/» _ !jl< 
When this value of a> 2 is substituted into Eq. 9.19, K A is 

= (GJQ-[ J/- - l]"' 8 (9<21) 



to 



# 



A plot of K A as a function of n for a given value of GB and eo^ is given in 
Fig. 9.2. It is possible to extend the gain-bandwidth product of vacuum 
tube amplifiers and some transistor amplifiers through the use of com- 
pensated circuits. 1 However, in the interest of compactness, this treatment 
is not included in this edition. 

PROB. 9.4. If GB = 409 x 10 6 radians/sec, and f H = 12 MHz, calculate K A 
for 3, 7, and 8 cascaded stages. Determine the bandwidth per stage (/a) and 
gain per stage for each amplifier. Answer: For n = 7,K A = J36,f 2 = 32 MHz, 
K = 2.03. 

When n identical singly tuned (or double tuned with critical coupling) 
amplifiers are cascaded, it can be shown 2 that 

B n = B(2 lln - 1)* (9.22) 

•Compensated circuits are treated in the following texts: Electronic Circuits by 
T. L. Martin, Prentice-Hall, Englewood Cliffs, N.J., pp. 123-149 and 169-171, 1955. 
Pulse and Digital Circuits by J. Millman and H. Taub, McGraw-Hill, New York, 
pp. 67-85, 1956. Fundamentals of Television Engineering by G. M. Glasford, McGraw- 
Hill, New York, pp. 163-197, 1955. 

• For derivations of this equation see Electronic Circuits by T. L. Martin, Prentice- 
Hall, Englewood Cliffs, N.J., pp. 171-174, 1955. 



342 Electronic Engineering 

where B n is the bandwidth of the total amplifier and B is the bandwidth 
of each stage in the amplifier. The behavior of these tuned circuits is 
much the same as the behavior of the jR-Ccoupled amplifier just considered. 
In fact, Eq. 9.22 indicates the bandwidth per stage must increase with an 
increasing number of stages if a given total amplifier bandwidth is to be 
maintained. Thus, an optimum gain exists for a given transistor or tube 
and a given bandwidth. 



35 



i — i — r 



i i — i — i — r 

GB = 314 xlO 6 
W H = 2tt (10 7 ) CPS 




6 7 8 9 
Number of stages 

Fig. 9.2. A plot of Eq. 9.21. 

Tuned circuits do have a rather unique advantage if the cascaded 
stages are not tuned to the same resonant frequencies. Tuned amplifiers 
of this type are known as stagger-tuned amplifiers. Again, the treatment 
of these amplifiers is deleted in the interest of compactness. However, 
excellent and detailed treatments exist in the literature. 3 

So far, the work on cascaded amplifiers has been primarily concerned 
with gain and bandwidth of the amplifier. If a step function is applied 
to a single stage amplifier, the rise time was given in Chapter 7 as being 
approximately equal to 2.2/» 2 . If this same concept is to be applied to 

3 The design of stagger-tuned circuits will not be included here. However, excellent 
derivations exist in Vacuum Tube Amplifiers by G. E. Valley and Wallman, Vol. 18 of 
the MIT Radiation Lab series, McGraw-Hill Book Co., New York, pp. 176-200, 1948, 
or Electronic Circuits by T. L. Martin, Prentice-Hall, Inc., Englewood Cliffs, N.J., 
pp. 186-205, 1955. 



Cascaded Amplifiers ^43 

cascaded stage amplifier, the expected rise time will be about 2.2jw H . 
In addition, Martin 4 has used a work by Elmore 5 to show that the overall 
rise time of the cascaded amplifier T AR is 

T AR = (7V + TV + TV + • • 0* (9-23) 

where 

T m is the rise time of stage one 
T m is the rise time of stage two, etc. 

Thus, if n identical stages with a rise time of T BS are cascaded, the rise 
time of the total amplifier T AB will be 

Tab = T m {n)* (9.24) 

These two equations assume no overshoot in the rise waveform and are 
accurate to within 10 per cent when as few as two stages are involved. 

PROB. 9.5. Design a video amplifier with a bandwidth of 30 Hz to 4 MHz. 
The total voltage gain must be at least 1000. Use 6AU6 tubes and assume the 
following data applies : 

C, = 5.5 pf 

C gv = 0.003 pf 
C = 5 pf 
g m = 5200 yumhos 
max R a = 1 megohm 
r„ = 1 megohm 
Assume wiring capacitance per stage is 3 pf. 

(a) Determine the figure of merit for one stage. 

(b) Determine how many stages are required. 

(c) Determine <a x and a> 2 . 
id) Determine R L and C c . 

(e) Determine R k , R s „ C sg , and C k (see the 6AU6 curves in Appendix III). 

(/) Draw the diagram for the total amplifier. List all values. 
PROB. 9.6. Design a video amplifier with a bandwidth of 30 Hz to 4 MHz. 
The total current gain must be at least 1000. Use 2N384 transistors (charac- 
teristics are given in Appendix III) and assume the interstage wiring capac- 
itance is 3 pf. 

9.2 db GAIN 

Power gain in bel units is defined as 

b = log- 2 (9.25) 

Pi 
* Op. cit., Martin, pp. 222-224. 

5 W. C. Elmore, "Transient Response of Damped Linear Networks with Particular 
Regard to Wideband Amplifiers," /. Appl. Phys., Vol. 19, pp. 55-62, January 1948. 



34* Electronic Engineering 

where "log" is the logarithm to the base 10 and b is in bels. PJP 1 is the 
power ratio between the points in question. 

The bel unit is convenient because it reduces a multiplication problem 
in the case of the gain of a cascaded amplifier to an addition problem. 
Nevertheless, the bel is an inconveniently large unit because a power 
gain of 10 is only 1 bel. Therefore, the decibel (db) has been accepted as the 
practical unit. The db unit has the additional advantage that a power 
change of 1 db in an audio system is barely discernible to the ear, which 
has a logarithmic response to intensity changes. 

db=101og£- 2 (9.26) 

P i 
Also 



db=101og^- 2 (9.27) 



' V 1 '/R 1 
If the resistance is the same at the two points of reference, 

db = 10 log (^J = 20 log - 2 (9.28) 

Similarly, 

db = 201og- (9.29) 

The impedance levels are frequently of secondary importance in a voltage 
or current amplifier. Therefore, the Eqs. 9.28 and 9.29 are sometimes 
loosely used without regard to the relative resistance levels. 

PROB. 9.7. An amplifier consists of four stages, each of which has a voltage 
gain of 20. What is the db gain of each stage? What is the total gain in db? 
Assume that the resistance levels of each stage are the same. 
PROB. 9.8. What is the db level at the half-power frequencies, / x and / 2 , 
compared with the mid-frequency or reference level? 

It is often convenient to express a power level in db with regard to a 
certain reference level. One commonly used reference level is 6 mw. 
Another commonly used reference level is 1 mw. When this (1 mw) 
reference level is used, the db units are usually called "volume units" 
(vu). On the other hand, the open-circuit output voltage of a microphone 
is usually rated in db with reference to 1 v when the standard excess 
acoustical pressure is 1 micro bar, or one millionth of standard barometric 
pressure. 

PROB. 9.9. What is the power level of 40 db? 40 vu? Answer: 60 w, 10 w. 
PROB. 9.10. What is the open-circuit output voltage of a microphone which 
has —56 db level? Assume the excess acoustical pressure to be 1 micro bar. 



Cascaded Amplifiers 345 

9.3 STRAIGHT LINE APPROXIMATIONS OF GAIN AND 
PHASE CHARACTERISTICS (BODE PLOTS) 

The analysis given in Section 9.1 is sufficient for the analysis and design 
of identical cascaded stages. This section will present a method of 
analysis which can be used on any amplifier. However, as an introduction, 
we will apply this method to a single R-C coupled stage. The transfer 
function of a single R-C stage was given by Eq. 7.56 as 

G K ^ (7.56) 

(s + w^s + eo 2 ) 

For a-c steady state, s = jo. Now, if co is assumed to be small in com- 
parison with m u the term (s + aij becomes essentially m v In addition, 
since a) 1 « co 2 in a typical amplifier, the term (s + cd^) becomes essentially 
a) 2 . Then, Eq. 7.56 becomes 

G=-K J — (9.30) 

Thus, the gain of the amplifier is proportional to frequency and the 
phase angle of G is 270°. An interesting method of expressing the magnitude 
relationship exists. In musical terms, the fre quency doubles every octav e. 
Thus, from Eq. 9.30, the gain doubles for every octave increase in fre- 
q uency . When Eq. 9.28 or 9.29 is used. 20 log 2 is approximately six. 
Thus, we can also state the gain increases 6 db per octave frequency increase. 
A plot ot do gain vs frequency over the range where Eq. 9.30 applies will 
be a straight line if frequency is plotted on a logarithmic scale. As noted, 
Eq. 9.30 is valid for a> « m x . However, as co approaches %, the accuracy 
of this approximation decreases. Nevertheless, as an approximation let us 
assume that Eq. 9.30 is valid for to <C co x . Then, the plot of G vs co and 
phase angle vs co will be as shown (for to < co^ in Fig. 9.3. 

Now, if s = jto and w 1 « to « co 2 , then (s + to x ) *»yto and (s + co 2 ) *** 
to 2 . Then, Eq. 7.56 becomes 

G= -K (9.31) 

In this case, the gain is a constant and is independent of frequency while 
the phase angle remains constant at 180. If this condition is assumed to 
exist (again, this is a rough approximation) for to x > co > co 2 , the plots 
will have the form given in Fig. 9.3. 

Finally, if s =ja> and a> » a> 2 , then, (s + coj) mjm and (s + co^ <=tsj<o. 
Under these conditions, Eq. 7.56 becomes 

G=-K^ (9.32) 



346 



Electronic Engineering 



In this case, G is reduced by one-half for each octave frequency increase 
or G decreases by 6 db per octave frequency increase. In addition, the 
phase angle becomes 90°. If these conditions are assumed to be present 
for a ^ co 2 , the plots will be as shown (for co > a> 2 ) in Fig. 9.3. 

The actual plots of gain and frequency (Fig. 7.16) are plotted as dashed 
lines in Fig. 9.3. The straight-line approximations are seen to be fairly 
good approximations. In fact, with a correction factor which can be 



III III 

G = ~K of ^ — Straight line approximation „^-G = 


-K 


V^ A 


A* — Actual curve \ 




/* Slope = 6 db/octave Slope = -6 db/octave "' 

/III III 





O.loii 



£01 



10o>i 



0.1C02 



0)2 



10O2 



(a) 



ob 270 



-1,180 



90 





1 1 1 


i 


1 


- 


- 




^^-^- Actual curve 


1 


i 




\ 


1 


1 





O.lcoi coi lOcoi 



0.1o)2 0>2 10o)2 



(b) 



Fig. 9.3. Straight-line approximations for G = —K[sco 2 l(s + coi)(5 + &> 2 )]. (a) Gain; 
(b) phase shift. 



applied near the singularities, accurate plots can be obtained. With the 
insight gained from the straight-line analysis of the R-C coupled stage, 
we are now ready to derive some general rules which can be applied in 
plotting gain and phase curves. 
The total amplifier gain G A was given in Eq. 9.4 as 



G A = G x x G 2 x G 3 x 



(9.4) 



In general, the stage gains (G u G 2 , etc.) are functions of s. Consequently, 
G A is also a function of s and will have the form 



G A = // 



s n + a^- 1 + a 2 s'- 2 + • • • a n 
s m + b^™- 1 -f b 2 s m ~ l + ■■■b m 



(9.33) 



Cascaded Amplifiers 34' 

In actual circuits, any number of the coefficients a x , b u a 2 , b 2 , etc. may be 
zero. The polynomial in the numerator and the polynomial in the 
denominator can be factored. When these polynomials are factored, the 
terms will have the following three forms: 

Form 1 = s k (9.34) 

Form 2 = (s + af (9.35) 

Form 3 = (a 2 + 2£co n s + co n *) k (9.36) 

where k is an integer 1, 2, 3, etc. which differs from one if repeated roots 
are present. Thus, Eq. 9.33 can be factored (in fact, since G A is usually 
written as the product of the individual stage gains, the terms are usually 
already factored) and written as 

GA = Hs As + Zl)(S + Z * y " (9-37) 

where h can have a value of zero, a positive integer, or a negative integer, 
and Z 1; Z 2 , P u P 2 , etc. may be complex numbers if they are derived from 
terms such as those of Eq. 9.36. Now, if we take the logarithm 6 of Eq. 
9.37 we have 

log G A = log H + h log (a) + log (s + ZO + log (s + Z 2 ) + • • • 

- log (s + P 1 ) - log (s + /»,) - • • • (9.38) 

Thus, if we work with logarithms of the different factors in Eq. 9.37, the 
responses of the different factors can be added or subtracted to obtain 
the response of the total amplifier. 

The log H is a constant and is independent of frequency. However, for 
the steady-state solution, a becomes ju> and 



h log (s) = h 



log <o +j- 



(9.39) 



As already noted, the term log co is equal to a gain increase of 6 db/octave. 
The effect of h log (s) on G A is plotted in Fig. 9.4 for several values of h. 
When considering factors of the form given by Eq. 9.35, it is convenient 
to draw, first of all, a straight-line approximation for the gain and phase 
characteristics and then correct this approximation in order to arrive at 
the actual gain and phase curves. Thus, we note that when s =jm, Eq. 
9.35 can be written as 

log [O + a)'] = k log (Jm + a) (9.40) 

6 In taking the logarithm of a complex number, the complex number should be 
reduced to a magnitude and a phase angle such as M». Then, log M» = log M +jd 
where is in radians. 



348 



o) 



Electronic Engineering 

h = 2 (Slope = 12 db/octave) 

ft = 1 (Slope = 6 db/octave) 
h = (Slope = db/octave) 




h = - 1 (Slope = -6 db/octave) 



V A = -2 (Slope = -12 db/octave) 
h = 2 



h=l 



h = 



h= -1 



h= -2 



(A> 
Fig. 9.4. A plot of A log (s) for several values of h. (a) Gain plot, (6) phase angle plot. 

If co « a, Eq. 9.40 reduces to (k log a) which is a constant. Thus, for a 
straight-line approximation, the gain and phase curves will be as shown 
in Fig. 9.5 (by the dashed lines) for values below co = a. However, if 



co » a, Eq. 9.40 becomes k 



log co +j- 



. In this case, the slope is 



(k X 6 db/octave) and the phase angle is — . For the straight-line 

approximation, these curves hold for co > a (Fig. 9.5). The two straight 
lines for the gain curve must intercept at co = a. When actual values 



Cascaded Amplifiers 



n Log a 

c 






a 



-Logo 



Actual curve 
ldb 



349 



""^•^ Straight line approximation 



■2db 
Slope = 6 db/octave 



0.5a 0.7a a 1.5a 2a 



>i.^- Slope = -6 db/octave 




Straight line approximation 
A=+l 




Fig. 9.5. A plot of k log (jot + a) for k = +1 and -1. (a) Gain; (6) phase angle. 

near to = a are calculated, the curves are as shown by the solid lines. 
Table 9.1 lists the actual error between the straight-line approximation 
and the actual curves for a few significant values of co. Note that k will 
be positive if the term (s + af is in the numerator of Eq. 9.37 and will be 
negative if the term is in the denominator. 

When conjugate poles or zeros exist (as in tuned amplifiers) the factors 
have the form given by Eq. 9.36. Again, a straight-line approximation 
can be used to simplify the plotting procedure. When s =jo>, Eq. 9.36 



350 Electronic Engineering 

can be written as 

log (-co 2 + 2j£a> n a> + cojf = k log (-co 2 + 2/fro n a> + w„ 2 ) (9.41) 

Now, if co « co n , Eq. 9.41 becomes 2k log co n which is a constant. This 
straight-line approximation is used for all values of co < co n (Fig. 9.6). 
In contrast, if co » co n , Eq. 9.41 becomes 2k log w + AyV. Note that if 
k is one, the slope of the straight-line approximation will be 12 db/octave. 

TABLE 9.1 
Correction Factor for Log [(/co + a) k ] 
Magnitude Correction To Phase Angle Correction 
co Be Added to the Linear To Be Added to the 

Approximation Linear Approximation 



0.5« 


k x 1 db 


0.707a 


k x 2db 


a 


k x 3db 


1.414a 


k x 2db 


la 


A: x 1 db 



A: 


X 


26.6° 


A: 


X 


35.3° 


/t 


X 


45° 


-A: 


X 


35.3° 


-A- 


X 


26.6° 



The interception of the two straight-line gain curves occurs at co = co n . 
The straight-line approximations and the actual curve for a typical case 
is shown in Fig. 9.6. However, since Eq. 9.41 contains £ as well as co n , 
the exact shape of the curve near co n is a function of £. Curves which can 
be used to determine the proper correction for different values of £ are 
given in Fig. 9.7. The corrections listed by these curves should be multi- 
plied by k to obtain the exact corrections to be used. The curves extend 
only to cojco n = 1.0. However, the attenuation curves are also valid for 
cojco when co > co n . With a change of sign, the phase shift curves are 
also valid for cojco when co > co n . 

The concepts just developed will now be used in an example to clarify 
their use. 

Example 9.1. The gain equation for an amplifier is given by the following 
equation : 

s 2 
A = l (s + 100)(j + 200X5 + 100,000) 2 (9 ' 42) 

Plot the amplitude and phase of G A as a function of frequency. 

As a first step, find the gain at some frequency which can be used as a reference. 
For example, assume s =/co =yl0. When this value is substituted into Eq. 
9.42, i]t becomes 

-100 

A ~ (yio + loox/io + 200X/10 + ioo,ooo) 2 (9-43) 



Cascaded Amplifiers 



351 



Slope (12 db/octave) 




Slope (-12 db/octave) 



pin u 



Straight line approximation 




Fig. 9.6. A plot of k log (-to 2 + 2/£cu„co + a>J) for k = +1 and k = -1. (a) Magni- 
tude ; (6) phase angle. 



352 



Electronic Engineering 



-5 



-10 



-20 



100° 

3 80' 

8P 
| 60- 

DO 

!§40« 
Si 
£ 20" 




f =2.0 



- 
















.*■ 


i 


- 






0.75 v 

°-5^ 




*< 


^ 


/, 


I 


- 










N 


f 


a; 




/- 


- 






^ 




^ 




^ 


/ 


















^f = 0.2 



0.1 



0.2 



0.3 0.4 0.5 0.6 0.7 0.8 1.0 

"Mi *- 

(b) 
Fig. 9.7. Magnitude and phase of (s* + 2fco„.s + co n a )/co n 8 . (a) Magnitude, (b) phase. 

This is approximately equal to G A = 0.5 x 10~ 8 . We can assume this is our 
reference level at <u = 10. 

As a next step, the straight-line approximations can be drawn. Accordingly, 
the individual straight-line approximations will be as shown by the dashed lines 
in Fig. 9.8. When these individual lines are added, the solid line (which is the 
straight-line approximation for G A ) results. 

In the final step, a correction table is tabulated as shown on page 355. 

When these total corrections are applied to the straight-line approximation 
(shown dashed in Fig. 9.9), the actual gain and phase curves shown as solid lines 
in Fig. 9.8 result. These are the required curves. 

The foregoing example indicates the process to be used in plotting gain 
and phase characteristics for any amplifier whose characteristic equation 



Cascaded Amplifiers 



353 




V 



The sum of all 
the lines = G A 



Line due to 

j^oirterm 

Slope = -6db/octave 

Line due to 
1 



.+200 lerm 
Slope — — edb/octave 



10 



Slope = - 12 db/octave 



100 200 500 1000 10,000 

oi in radians/second. 




100,000 



1,000,000 



•E 



10 



"1^- 



Phase shift due to s 2 term 



-Total phase shift 



J I 



-J_ _ 

100 200 500 1000 



Phase shift due 

to .-+W ,erm "1 



I 



10,000 100,000 

u in radians/sec - 



1,000,000 



I 
.x. 



Phase shift due to 
HfW ,erm 



Phase shift due to 
f.+W 1 *"""" 



Fig. 9.8. The straight-line approximations for Eq. 9.42. 



354 



Electronic Engineering 



50 



40 



30 



20 



10 


-10 











\ 
\ 




>ht line approxir 
gain curve 






Actual 















































10 



10 



100 



1000 10,000 

b> in radians/sec 



100 



1000 10,000 

to in radians/sec 



100,000 



100,000 



1,000,000 





1 










^"■~~»». 












>v | Straight line a 
\l 


jproximation 






V 


> ^ 








2. 




\^---ActL 
iV 


al curve 






o 




1 \^ 

1 -■«« 
i 








IT 












■I 

























1,000,000 



Fig. 9.9. The actual gain and phase curves for Eq. 9.42. 



Cascaded Amplifiers 



355 



TABLE 9.2 

Correction Values to be Used on Fig. 9.8 

Correction for Correction for Correction for 
1 . 1 1 



Frequency s + ioo 



term 



s +200 



.term 



(s + 100.000) 2 



Total 
Correction 



db Phase db Phase db Phase db Phase 



50 


-1 


-26.6° 










-1 


-26.6° 


70 


-2 


-35.3" 










-2 


-35.3° 


100 


-3 


45.0° 


-1 


-26.6° 






-4 


+ 18.4° 


140 


-2 


+35.3° 


-2 


-35.3° 






-4 


0° 


200 


-1 


+26.6° 


-3 


45.0° 






-4 


+71.6° 


280 






-2 


35.3° 






-2 


+35.3° 


400 






-1 


+26.6° 






-1 


+26.6° 


50,000 










-2 


-53.2 


-2 


-53.2° 


70,700 










-4 


-70.6° 


-4 


-70.6° 


100,000 










-6 


-90.0° 


-6 


-90.0° 


141,400 










-4 


+70.6° 


-4 


+70.6° 


200,000 










-2 


+53.2° 


-2 


+53.2° 



is known. In fact, it is often possible to start from a known gain curve 
and synthesize an amplifier which will have this gain curve. 

PROB. 9.11. (a) Draw the straight-line approximation curves (gain and phase) 
for an amplifier whose gain is given by the equation 



° A ~ (s + 100)0 + 1000)V + 6 x 10*s + 10 10 )0 + 10 5 ) 
(b) Draw the actual gain phase curves for this amplifier. 
PROB. 9.12. Repeat Prob. 9.1 1 if the gain equation is 

s(s + 200) 2 



G, = 



is + 100)0 + 400)0 + 800)0 + 10 6 ) 3 



PROB. 9.13. Three identical stages are cascaded. The gain equation for one 

stage is 

s 



G. = 



is + 100)0 + io 6 ) 



(a) Plot gain vs frequency curves for the total amplifier. From this curve, 
find a> L and u> H for the total amplifier. 

(/>) Use the method outlined in Section 9.1 to determine a> L and co H . How 
do these values compare with those in part a ? 



356 



Electronic Engineering 



PROB. 9.14. It is desirable to construct an amplifier which has a gain vs 
frequency curve with the same shape as the one given in Fig. 9.10. 

(a) Find a gain function G A which would give this type of response. 

(b) Break this function up into functions which represent R-C coupled stages. 

(c) Transistors are available which have the following characteristics: 

h ie = 2000 ohms 

h re <*> 

h„ = 50 

h oe = 2 /wnhos 

+6 



-6 



-12 

-o 
-o 

!-18 



-24 



-30 



-36 

-42 

31 62 125 250 500 IK 2K 4K 8K 16K 32K 64K 256K 1M 2M 4M 8M 16M 
10 in radians/sec 

Fig. 9.10. Characteristics for Prob. 9.14. e 

If the total shunt and wiring capacitance per stage is assumed to be 250 pf 
and p cutoff is assumed to be muchgreater than co^design the amplifiers which 
will produce the required response. Answer: (b)G A = [K x sl(s + 125)(s + 2 p. 
70 8 )] x [Ktsl(s + 550)0* + 7 x Kfi)] x [K 3 sl(s + 350)(s + 7 x l(fi)]. 































• 






















































1 , 


S 




















\ 
\ 
















1 / 






















\ \ 
















/ 






















V ' 
















! 
























i 












1 


























k\ 


















































ll 


























\\ 












ll 


























\\ 










1 


'/ 


























y 


\ 






i 


t 




































1 







































9.4 CASCADING INTERACTING STAGES 

In most transistor and in some high-frequency tube applications, the 
input impedance of one stage depends on the load of this stage. In fact, 
as found in Chapter 5, the input impedance of a common emitter amplifier 
was given as 

R, = h,. - Kehu (5.16) 



Ke. + g L 



Y o - h o - r^rV ( 9 - 45 > 



Cascaded Amplifiers S 57 

This equation was developed for a common emitter circuit, but the 
solution is actually general and applies to the A parameters in any 
configuration. Thus, the input impedance can be written as 

z < = h < ~ -TTV {9M) 

When stages are cascaded, the Y L of one stage includes the input impedance 
to the next stage as well as the coupling network. Thus, Y L will usually 
be a function of s and, if many stages are present, may be a rather complex 
function of s. In addition, the output admittance of a stage is given by 
the relationship 

M/ 

K + z 

where Z a Js the impedance of the driving generator. From this relation- 
ship, we note that we not only have interaction from the output stages 
of a cascaded amplifier back toward the input stages, but also the input 
stages will interact with the output stages. 

Obviously, the design of even a three- or four-stage amplifier can be 
very complex unless something is done to reduce the effects of the inter- 
action. Two methods of reducing the effects of interaction have already 
been discussed in Chapter 8. The first of these methods (Section 8.5) was 
to neutralize each stage. When a stage is neutralized, h r is reduced to 
zero. Under these conditions, Z, = h^ (also Z = h ) and the stages are 
completely noninteracting. Unfortunately, it may be difficult or in- 
convenient to neutralize a stage completely (especially over a wide range 
of frequencies), since the magnitude and phase angle of h r are both 
frequency sensitive. 

The second method of reducing interaction is to deliberately mismatch 
the load 7 as outlined in Section 8.6. Note in Eq. 9.44 that if Y L « h oe , 
the second term on the right side becomes p*shji 1 \h and a change of Y L 
will have very little effect on Z t . In the other extreme, if Y L becomes 
quite large, the second term on the right of Eq. 9.44 can be made so much 
smaller than h ( that Z t becomes essentially equal to h t . In order to be 
more quantitative, let us define an interacting parameter d t as the fractional 
change of Z< with a change of Y L . 

h r h, 

= dzjdY L = (h + Y L f Yl (946) 

' 2JY L h h r h f 

' K + Y L 
' J. F. Gibbons, "The Design of Alignable Transistor Amplifiers," Technical Report 
106, Stanford Electronics Laboratories, May 7, 1956. 



358 Electronic Engineering 

A straightforward reduction of this equation yields 

6 t = ^^ (9.47) 

[hlK + Y L ) - h r h f ](h + Y L ) 

Now, if Y L is much greater than h , the term (h + Y L ) an Y L . Then, 
Eq. 9.47 can be written as 



Kh f Y L = h r h, 
h t Yj? - h r h,Y L h t Y L - h r h f 



(9.48) 



The interaction between stages is reduced as d ( is reduced. For example, 
if d { is equal to 0.1, a 10 per cent change of load impedance is reflected 
back to the input as a 1 per cent (0.1 x 10 per cent) change of input 
impedance. Thus, for low interaction, 8 i should be small. Now, if <5 4 is 
much less than unity, h t Y L » hjt f in Eq. 9.48. If this is the case, the 
equation for d t becomes 



& 



h J± (9.49) 

h t Y T . 



For the other extreme, Y L will be much less than h . In this case, the 
term (h + Y L ) «ss h and Eq. 9.47 can be written as 

A _ — M^ — (9 5Q) 



{KK - h T h t )h 

For a given value of d ( , the appropriate value of Y L can be determined 
from either Eqs. 9.49 or 9.50. The exact equation to use will be determined 
by the circuit configuration. For example, the value of Z t is usually much 
less than Z„ in a common-emitter amplifier. Therefore, it is usually 
desirable to make Y L > h oe and Eq. 9.49 would be applied if common- 
emitter stages are used. On the other hand, if an emitter follower is to be 
used to drive a high input impedance circuit, it would be desirable to let 
h oe > Y L , and Eq. 9.50 would be used. 

The problem of interaction between stages is usually more pronounced 
in tuned amplifiers. In fact, just aligning a tuned-multistage-interacting 
amplifier can be a very tedious task. Of course, designing such an 
amplifier can be complex and difficult. Consequently, an example in the 
design of a multistage amplifier will be given. 

Example 9.2. A three-stage I.F. amplifier is to be designed as shown in Fig. 9. 1 1 . 
The desired center frequency is 500 KHz and a bandwidth of 10 KHz is required. 
Determine all pertinent parameters in this circuit if d t is to be restricted to a 



Cascaded Amplifiers 



359 



magnitude of 0.1. The transistor has the following measured parameters at 
500 KHz: 

h ie = iooo a 
h n =yio- 3 

*/. = 25 

h m = (8 +y30)10-« = (31 x lO" 6 )/^ mhos 

First, let us determine the required magnitude of Y L from Eq. 9.49. 



\Y L \ = 



h t 8 



= 25 x 10 -5 mhos 



(9.51) 



In order to determine the real and reactive components of Y L , a knowledge of 
the output admittance of the transistor must be known. The output admittance 
can be calculated from Eq. 9.45 if the total generator impedance is known. If 




Fig. 9.11. An a-c circuit of a three-stage I.F. amplifier. (The d-c bias circuits are not 
shown.) 

we assume the coupling circuit is adjusted so the input of the transistor is matched, 
the generator impedance of Eq. 9.45 is equal to Z t of Eq. 9.44. When this sub- 
stitution is made, Y is equal to 

Y n =K **£__ (9.52) 



hi +hi - 



K + Yr 



Then 



y = (8 + /30)10- 



25 x lO" 3 ^*^ 



2 x 10 3 - / 



25 x 10- 3 



(8 + /30)10- 6 + (25 x 10- 5 K? 



While the phase angle associated with Y L is not known, the effect of Y L on Y 
is essentially negligible in this example. (In fact, the effect of Y L becomes smaller 
as 8i becomes smaller.) If the effect of Y L is not negligible, assume a value for Y L 
and substitute this value in Eq. 9.52. Use the value of Y so obtained to determine 
a tentative value for Y L . Now use the value of Y obtained from this tentative 
value of Y L in Eq. 9.52 to arrive at a final value for Y L . The value of Y for this 
example is 

Y * (8 + /17.5) x 10- 6 (9.53) 



360 Electronic Engineering 

Maximum transfer of energy from a generator with an admittance of G +jB 
is possible when the load has an admittance of G — jB. In this example, the 
magnitude of Y L must be 250 micromhos. However, maximum power will be 
delivered to the load if the reactive term of Y is cancelled by the load admittance. 
Thus, Y L =(249 -;17.5) x lO" 6 or 250 x 10- 6 Z=«£mhos. 

The desired bandwidth for this amplifier is 10 KHz. However, since there 
are three stages, Eq. 9.22 is used to find the required bandwidth per stage : 

B n 10 KHz 

B = (2 i/» _ dx = (2 >A - i>K = 196 KHz < 9 - 54 > 

From Eq. 8.50, the Q of the output tuned circuit must be 

a> 500 KHz 
G =¥ = lT6KH-z= 25 - 5 (955) 

Now, if the unloaded Q of L x and Q in Fig. 9.1 1 is much higher than 25, essen- 
tially all of the loading on the output tuned circuit is due to the load resistance R L . 
Therefore 

** - Wy7) - 249-lTKP - 400 ° ° hmS (956) 

From Eq. 8.9, the value of C t can be found. 

Q 25.5 

C, = — = - A 77S t^ ? 77T, = 0.00203 /*fd (9.57) 

1 Ro> 4 x 10 3 x 6.28 x 5 x 10 5 

The admittance of C ± at co is y'6.28 x 5 x 10 5 x 2.03 x 10~ 9 or 6.37 x 10~ 3 
mhos. The desired reactive component for Y L is — J 17. 5 x 10 -6 . Therefore, the 
total admittance of L x is -y6370 x 10" 6 -y'17.5 x 10" 6 = -y'6.387 x 10~ 3 . 
Thus, 

^ = ^ = 6.28 x 5 x 10* x 6.387 x 10- = 499 " taL (958) 

With the output circuit designed, we are now ready to consider the interstage 
coupling network L 2 , L3, and C 2 . The input impedance to the output transistor 
(from Eq. 9.44) is 

/10- 3 x 25 

Zi = iO 3 ~ 7Z n^T^=5 7^ .,-,*„« e *> 1000 - J10O (9.59) 

* (8 +y30)10-* + (249 -yi7.5)10' 6 

The output admittance of transistor T 2 is the same as the output admittance of 
transistor T 3 . Thus Y =;(8 +/17.5) x 10~ 6 . The desired load on T 2 will be 
the same as the load on T 3 or (249 — y'17.5) x 10 -6 mhos. 

Hence, as noted in Eq. 9.56, the equivalent parallel resistance R v for the circuit 
C 2 L 3 would be 4000 ohms. Using Eq. 9.57, the value of C 2 will be equal to 
0.00203 /*fds. The desired loaded admittance of the coil Z, 3 is given by Eq. 8.80 as 

(<o Mf 
Z in = R L3 +/-A + Ru+Zi+JaLt (?»■«) 



Cascaded Amplifiers 361 

Now Z t is over 1000 ohms, so it is much larger than R L2 or R L3 . In addition, 
ja>L 2 can be used to cancel the -y'l 00 component (see Eq. 9.59) of Z v Then 

100 

^ = 6.28x5x# =31 ^ henrieS (961) 

Under these conditions, L 3 furnishes all of the reactance in the primary circuit. 
The admittance of L 3 must be equal to the admittance of C 2 which is — /6370 
^mhos plus the — jl7.5 /xmhos of the required Y . Thus, L 3 must have the same 
admittance as Z,j. Hence 

L 3 = 49.9 n henries (9.62) 

The value of M can be determined by using Eq. 8.83. For this example 

/L 3 \* Z49.9 x 10- 6 V 
R v = Ml Re(R t ) = 4000 = I — j 1000 (9.63) 

or 

49.9 
M x = —r- =25/* henries 

The value of the coupling coefficient k x will be 

Mj _ 25 x 1Q- 6 _ 

kl = (Z^JW = (31.8 x 49.9) x lO" 6 - ° 63 (964) 

All of the components in the second coupling circuit have now been determined. 
The coupling circuit between T x and T 2 is coupling exactly the same imped- 
ances as the coupling circuit between T 2 and T 3 . Consequently, C 3 = C 2 , 
L b = L 3 , L 4 = L 2 , M 2 = A/j, and k 3 = k x . 

The foregoing example used impedance mismatching as the means of 
reducing interaction between stages. Of course, other interstage con- 
nections and other approaches 8 to the problem are also used. When 
impedance mismatching is used, the total power gain is reduced below 
that which can be obtained with a properly matched and neutralized 
stage. If too much power gain is lost through impedance mismatching 
and the circuit is extremely difficult to neutralize, a combination of the 
two processes may be used. A simple feedback circuit is employed to 
produce partial neutralization and impedance mismatching is employed 
to reduce the interaction to the desired level. 

PROB. 9.15. Determine the total power gain at eo = 500 KHz of the three- 
stage amplifier of Example 9.2. 

8 A graphical approach to the design of transistor (or tube circuits) is given in Chapters 
1 1 and 1 8 of Transistors and Active Circuits by J. G. Linvill-and J. F. Gibbons, McGraw- 
Hill, New York, 1961. 



362 



Electronic Engineering 



PROB. 9.16. Design a three-stage neutralized amplifier using the transistors of 
Example 9.2. The amplifier should be designed to produce maximum power 
gain and still have the same bandwidth and the same center frequency as in 
Example 9.2. Use inductively coupled circuits to couple the three stages. 

9.5 D-C AMPLIFIERS 

The need frequently arises for an amplifier that will amplify signals 
which vary so slowly that capacitors of practical size cannot be used for 
either coupling or bypass. Transformers are also useless in this application. 
Therefore, one amplifier must be directly coupled to another and the bias 
currents may not be separable from the signal currents. This type of 



oV. 




Fig. 9.12. One type of d-c tube amplifier. 



amplifier is called a direct coupled or d-c amplifier. Special problems 
arise in these amplifiers because of the restricted freedom in the choice of 
bias potentials and techniques. Transistors offer some advantages when 
compared with tubes because of the availability of both p-n-p and n-p-n 
types. However, the transistor thermal currents may present a serious 
problem in a high-gain d-c amplifier because the thermal currents may be 
amplified along with the signal currents and therefore the thermal currents, 
or their variations, must be kept small in comparison with the signal 
currents. Some typical d-c amplifier circuits will be discussed in this 
section to illustrate the special problems in d-c amplifier design and some 
techniques which may lead to satisfactory solutions of these problems. 

Although vacuum tubes are seldom used in present-day d-c amplifier 
design, this type of amplifier will be discussed here as an introduction to 
some of the problems encountered in d-c amplifiers. One type of tube 
amplifier is shown in Fig. 9.12. The first stage of this amplifier may have 
fairly high voltage gain even though the cathode bias resistor is unbypassed 



Cascaded Amplifiers 



363 



because the bias resistor R K1 may be small in comparison with the plate 
load resistor R PV The voltage gain is the same order of magnitude as 
r piI r ki because the output voltage is i P R P1 and the input voltage v t is the 
same order of magnitude as i P R KV However, a problem arises in the second 
amplifier K 2 because the grid of this amplifier is at the same d-c potential 
as the plate of V x . This high-positive grid potential with respect to ground 
requires an even higher positive cathode potential in order to provide 
negative grid bias. One way to provide this high-positive cathode potential 
is to use a large cathode resistor R K2 , but then the voltage gain of the 
amplifier V 2 is very small because R K2 is the same order of magnitude as 



£0. 




Fig. 9.13. A Darlington transistor circuit. 

R P2 and the input voltage v I2 = i P R K2 + v aK must be of the same order of 
magnitude as the output voltage. A bleeder resistor R b may bemused to 
increase the current through R K2 and therefore, to reduce its value. But 
in order to reduce the value of R K2 sufficiently to provide good voltage 
gain, the bleeder current must be large and the resulting heat generation 
and power supply diain may be serious disadvantages. A high-gain d-c 
amplifier may require several stages and since the plate potential of each 
tube must be of the order of 100 volts positive with respect to its grid, 
and hence the plate of the preceding tube, the supply voltage V PP may be 
very large in comparison with that required for an R-C coupled amplifier. 
Another disadvantage of the amplifier of Fig. 9.12 may be the large d-c 
bias voltage which is associated with the signal voltage in the output v . 
Batteries could be used to buck out the undesirable d-c potential rises, 
but batteries are bulky, produce a large shunt capacitance and habitually 
run down and require replacement. Gas diodes could be used instead of 
batteries, but these tubes generate noise which may mask the signal. 

The problems encountered in the tube amplifier may be minimized or 
avoided by the use of transistors. One commonly used transistor 
arrangement is shown in Fig. 9.13 and is known as the Darlington 



364 



Electronic Engineering 



connection. From the common-base configuration, 
isi = 'ei — 'ci = 'jElO — a l) 

'C2 = a 2'Bl == a 2'£l(l a l) 



Consequently, 
In transistor 2 
and 



lCT — ici + *C2 — ( a l + a 2 — a 2 a l) J £l 

Hence, the effective a of the entire circuit is 



l CT 



'■eft 



= («i + a 2 - a 2 ai) 



(9.65) 
(9.66) 
(9.67) 
(9.68) 

(9.69) 



To illustrate the effectiveness of the circuit, consider the case when 
a x = a 2 = 0.98. In this case, <x. eff is equal to 0.9996, which is quite close 
to 1. The circuit of Fig. 9.13 can be rearranged as shown in Fig. 9.14. 




oc 



Fig. 9.14. The "common-emitter" configuration of the circuit of Fig. 9.13. 



From Chapter 5, 



hfe 



1 -a 

the value of h fe for Fig. 9.14 (o^ = <x 2 = 0.98) would be 

0.9996 



(5.44) 



hfe = 



= 2499 



1 - 0.9996 

A single conventional common-emitter circuit containing a transistor 
with an a of 0.98 would have an h u of 49. (The actual gain of one con- 
ventional stage would be less than h fe .) Thus, if two conventional common- 
emitter stages are cascaded, the maximum possible gain would be h fe 2 . 
For the transistors with h u of 49, the maximum possible gain would 
accordingly be 2401. The Darlington configuration not only obtains a 
higher current gain, but also eliminates a number of circuit components. 



Cascaded Amplifiers 365 

In addition to the current gain characteristics, the input impedance 
characteristics of the Darlington connection are also of interest. For 
example, if the input impedance of transistor 1 is Z n , the input impedance 
of transistor 2 can be found from Eq. 7.130 of Chapter 7. y<\ It* ■ 

R in = h ie + (1 + h u )R L (7.13?) 

This equation was derived for a transistor with a load R L in the emitter 
circuit and R L « R d . These conditions are satisfied by transistor 2 of 
Fig. 9.14. The load R L in the emitter circuit of transistor 2 is Z a . 
Therefore, Eq. 7.132 becomes 

Z i2 = h M + (1+ h f JZ A (9.70) 

where the subscript 2 denotes the parameters of transistor 2. In effect, 
the input impedance has been increased by a factor h u . Of course, the 
value of load impedance connected to the output terminals will also 
influence Z i2 and even Z 41 as well as the total current gain. Nevertheless, 
the circuit of Fig. 9.14 is a very high current gain circuit and also has a 
high input impedance. 

The circuit given in Fig. 9.13. can be extended to include more than two 
transistors. Some manufactures include two or three separate transistors, 
connected in a Darlington configuration, in a single case. 

One problem that may arise in the Darlington connection results from 
the thermal current I co of transistor T t which is multiplied by the stability 
factor of T 2 , then enters the base of T lt and is again multiplied, this time 
by the current gain of T t . If germanium transistors are used, the com- 
ponent of thermal current in the output may be prohibitively large. 

The thermal current which reaches the output may be greatly reduced 
if the linear stabilization circuit of Fig. 9. 1 5a is used. Then parts of the 
thermal currents are shunted through the base resistors as discussed in 
Chapter 5 and are not amplified by the transistors. The stabilizing 
resistors reduce the gain of the amplifier because signal currents are also 
shunted through the base resistors. However, since the dynamic input 
resistance h ( of a transistor is much lower than its static input resistance 
h z over the useful operating range (Fig. 9.15ft), the thermal currents can 
be effectively reduced without a serious reduction of signal gain. 

The circuit of Fig. 9.16 uses reversed biased diodes to conduct the 
thermal currents. The signal currents are not shunted appreciably because 
the dynamic resistance of a reverse-biased diode is very high. The diodes 
should be chosen so that their reverse bias (saturation) currents are 
essentially equal to the transistor thermal current I co which should flow 
through its associated diode. Good thermal conduction should be main- 
tained between the transistor and the diode in its base circuit so com- 
pensation will be maintained over a wide ambient temperature range, and 



366 



Electronic Engineering 




(a) Circuit 




Base current 
(b) Operating point 



Fig. 9.15. A linearly stabilized Darlington-connected amplifier, (a) Circuit; (b) operat- 
ing point. 



with varying transistor dissipation. Positive, zero, or negative values of 
current stability factor may be obtained with diode stabilization, depending 
upon the characteristics of the diodes and the effectiveness of the thermal 
conduction. 

In some applications, the high input resistance may be a disadvantage 
of the Darlington connection. A very high input resistance, as well as 
very high current gain, may be obtained if three transistors are used in 
conjunction with thermal compensating diodes. 



•Rl 



€£ 



. 



Or 



Fig. 9.16. A diode-stabilized Darlington circuit. 



Cascaded Amplifiers 



367 



PROB. 9.17. A 2N1905 and a 2N1415 are used in the Darlington circuit shown 
in Fig. 9.15a. Let the base resistors R B draw half as much current as the 
respective bases at maximum current. The load resistance R L is 10 ohms and 
the supply voltage V cc = -20 v. Using R E = 0.5 ohm, determine suitable 
values for the other resistors in the circuit. Determine the magnitude of input 
current i z required to drive the output transistor into saturation. Determine 
the approximate input resistance of the circuit. Answer: R m = 4 KO, R B2 = 
150 n, i t = 0.425 ma, R in ~ 1.6 KCl. 

Another type of d-c amplifier is illustrated in Fig. 9.17. In this amplifier 
the transistor types alternate from n-p-n to p-n-p. Although only two 



<+V cc 



rm 



<5 



»o 




o + Vcc 



(a) 

Fig. 9.17. A d-c amplifier using n-p-n and p-n-p transistors, (a) Basic circuit; (6) 
stabilized and linearized circuit. 



stages are shown in Fig. 9.17, any desired number can be used. The 
collector current of the first amplifier is the base current of the second and 
so on, so the total current gain may be approximately equal to the product 
of the beta's of the transistors used. In contrast to the Darlington con- 
figuration, the input resistance of this amplifier is that of a single transistor 
and does not increase with the number of stages used. The thermal 
currents of each transistor are multiplied by all the transistors which 
follow. Therefore, at least the low-level stages should use silicon 
transistors. 

The thermal currents can be shunted through reverse-biased diodes 
(Fig. 9.17Z>). In the circuit given, diode D t should have a saturation 
current equal to the I col of transistor T x and similarly the saturation 
current of diode D 2 should match the I C02 of transistor T 2 . However, if 
the saturation current of diode D 2 were equal to the sum of I C02 and 
S I I col where S r is the current stability factor of the first stage, the diode 
D x could be eliminated. Then the thermal currents of both stages would 



368 



Electronic Engineering 



be shunted through D 2 . The main problem may be locating a diode with 
the desired saturation current. 

An alternate solution to the thermal stability problem is given in Fig. 
9.18. In this circuit, the stabilizing diode has been replaced by transistor 
T s . The thermal current which will flow through this transistor is SjI C03 , 
where S z is the current stability factor of the circuit of T a . But this 
stability factor can be adjusted over a wide range by varying the base 
circuit resistance R B , as discussed in Chapter 5. Therefore, resistor R B can 





o+V cc 



Fig. 9.18. A transistor thermal-stabilization circuit. 

be adjusted for optimum temperature compensation, the resistors R E1 
and R E2 are used to maintain fairly constant gain and input resistance 
over a wide range of input signal levels. 

PROB. 9.18. A 2N1905 and a 2N2712 transistor are used in the circuit of 
Fig. 9.\lb. The load resistance R L = 10 CI and the emitter resistor R E2 = 0.5 CI 
and V cc = 20 v. Determine a suitable value for R m and specify the desired 
saturation currents of diodes D 2 and D x at 25 °C. Determine the values of the 
input current // and the input voltage vj required to drive transistor T 2 into 
saturation. 

The final type of d-c amplifier to be considered is the differential 
amplifier shown in Fig. 9.19. There are two basic types, balanced and 
unbalanced. In the balanced amplifier, the input signal comes from a 
symmetrical source such as a strain gage, a bridge circuit, a balanced 
transmission line and so forth, where one side of the signal is not referenced 
to ground. With this type of signal, the forward bias on one base is 
increased while the forward bias on the other is decreased. If the two 
transistors are matched and linear, the increase of emitter current in the 
one transistor will equal the decrease in the. other and the total emitter 
current through R E will be constant. Therefore, R E , although very large, 
will not cause degeneration of the input signal since the emitter potential 



Cascaded Amplifiers 



369 



remains constant. In fact, if the transistors are not well matched, the 
emitter resistance R E will tend to degenerate the higher gain transistor 
and regenerate the lower gain transistor so that their respective emitter 
currents will match. Of course the collector potential of one transistor 
decreases while the collector potential of the other increases. The output 
voltage, which is the difference between the collector potentials, may be 
fed to a balanced load or to another balanced amplifier. The balanced 
amplifier which follows should use the opposite type transistors (p-n-p 
to follow n-p-n etc.) for easy attainment of proper bias potentials. 




o +V CC 



-o +V a 



t 




u 



t 

v 

I 

x 



'Re 



O-Vr, 



Vcc 



(a) W 

Fig. 9.19. A differential amplifier, (a) Balanced; (6) unbalanced. 

From the foregoing observations and inspection of Fig. 9.19a, we can 
see that in comparison with a single transistor (similarly biased and with 
zero emitter circuit resistance) the balanced amplifier provides: 

(a) The input resistance of the balanced amplifier is twice as high. 

(b) The output resistance of the balanced amplifier is twice as high. 

(c) The current and voltage gains are the same in both amplifiers. 

The main advantage of the balanced amplifier in a d-c application is 
that common mode input signals do not alter the output signal. That is, 
any signal which causes the collector currents of both transistors to 
increase or decrease equally and simultaneously will have no effect on 
the output voltage, which is the difference of the collector potentials. 
Therefore, changes in I co and V„ K owing to temperature change do not 
affect the output, providing that the transistors are matched and maintained 
at the same temperature. Thus thermal currents are not passed from stage 
to stage. In addition, the individual transistors have excellent thermal 
stability because the emitter resistance R E can be very large. Sometimes 



370 Electronic Engineering 

a transistor is used in place of R E to provide very high dynamic resistance 
while permitting the desired <2-point currents to flow. 

The circuit of Fig. 9.19a may be used to obtain an unbalanced output 
voltage, (or voltage with reference to ground) from either collector 
terminal. However, the output voltage is then reduced by a factor of two 
and the thermal component of collector current will not be cancelled in 
the output. However, the effect of variation of v BE with temperature 
will tend to cancel and the current stability factor of the amplifier may be 
very good (low) because of the large usable value of R E . 

The circuit of Fig. 9.196 has both unbalanced input and output. The 
base of one transistor is used as the ground reference, although any other 
point in the circuit could have been grounded, providing that the d-c 
voltage shift at the input could be tolerated. The base bias of transistor 
7\ is obtained through the driving source conductance. If the source does 
not have a conducting path, base bias resistors must be provided as shown 
in Fig. 9.19a. At first glance, the impression might be gained that the 
current through R E will not be constant in the unbalanced circuit and 
serious degeneration will occur. However, the dynamic emitter-base 
resistance (h ib2 ) of transistor T 2 is in parallel with R E . Since h ib2 is normally 
very much smaller than R E , the input resistance of the amplifier is, 
assuming R c « r d , 

^.= V + (*w+1)*m (9-71) 

But if the transistors are alike (h fel + \)h ib2 = h iel and 

R in ^ 2h M (9.72) 

Thus the input resistance of the unbalanced and balanced amplifiers is 
the same. 

A balanced output could be obtained from the circuit of Fig. 9.196 if 
the collector circuits were arranged as shown in Fig. 9.19a. Since h ib is 
very small in comparison with R E , Ai E2 ~ — A.i m and equal collector 
circuit resistors would produce approximately balanced voltages of 
opposite polarity. 

In the unbalanced circuit of Fig. 9. 196, the voltage and current gains 
are only one-half those of the balanced amplifier for the same collector 
circuit resistance because only one half of the available output is utilized. 
The thermal currents do not cancel in the output but the stability factor 
may be very good because of the large value of R E and small value 
of R B . As before, the effects of the thermal variation of V BE tend to 
cancel. For example, if V BE1 and V BE2 decrease by equal amounts, the 
potential of the base of T x with respect to ground does not change. 
Observe that the unbalanced amplifier of Fig. 9.196 is a common collector 
amplifier directly coupled to a common^base amplifier. 



Cascaded Amplifiers 37 1 

An index of goodness for the difference amplifier is the common mode 
rejection ratio 

Voltage Gain for Difference Signals 

CMRR = ■ —. — —r (9.73) 

Voltage Gain for Common Mode Signals 

Assuming that R c « 1/A„ so that G t ~ h u , then the voltage gain for 
difference signals in the balanced amplifier is 

G vD ~G£2R c l2h u )~h u R c lh ie (9.74) 

As previously discussed, the voltage gain of an amplifier with a large 
unbypassed emitter resistor (R E » r e ) is approximately R C IR E . In the 
difference amplifier the effect of R E is doubled because both emitter 
currents flow through R E . Therefore, the common mode gain is 

G v = 2RJ2R E = R C IR E (9.75) 

Substituting Eqs. 9.75 and 9.74 into Eq. 9.73. 

CMRR = ^^ (9.76) 

Transistor manufacturers provide two transistors in one case for use 
as difference amplifiers. In fact, these transistors are etched on a single 
silicon wafer so their characteristics will match and their temperatures 
will be as nearly the same as possible. Very low-drift difference amplifiers 
can be made from these units. 

PROB. 9.19. Two 2N2712 transistors are used in the unbalanced differential 
amplifier circuit of Fig. 9.196. The load resistor R c = 2.0 Kfl and the supply 
voltage, + V cc to - V cc , is 20 v. 

(a) Determine suitable values for the emitter resistor R E if V cc to ground 
is 10 volts. Answer: R E = 2 KS1. 

(b) Determine the approximate voltage gain, current gain, and current 
stability factor for the circuit. 

(c) What is the common mode rejection ratio of this amplifier? 

In addition to the foregoing d-c amplifier configurations, a circuit 
known as a "chopper" is also used. The crudest form of the chopper is 
shown in Fig. 9.20a. The switch S ly which may be either motor driven, a 
vibrating reed, or an electronic switch, is switched very rapidly between 
positions 1 and 2. Switch S x must operate much faster than the highest 
frequency of the incoming signal v v As a result, the output signal is 
transformed into a series of pulses whose amplitude is determined by v x 
(Fig. 9.206). These pulses are amplified by conventional a-c amplifiers 
and then "detected" or rectified (see Chapter 14) to restore the low- 
frequency or d-c signal. The chopper eliminates most of the stabilization 
problems mentioned but introduces some new problems. 

\ 



372 



Electronic Engineering 



\ 



v o 



(a) 




-Tl-nl, 



(b) 



Fig. 9.20. A very crude chopper circuit, (a) Chopper circuit; (6) voltage waveforms. 

The chopper circuits are beyond the scope of this text and consequently 
are introduced here so the reader will be aware of their existence. For a 
more detailed analysis the reader is referred to the literature. 



9.6 NOISE IN AMPLIFIERS 

Except in very broad band amplifiers where the available gain is 
determined by the gain-bandwidth product of the stage, the maximum 
useful power gain of an amplifier cascade is ultimately determined by the 
noise generated in the amplifier. For example, the question may arise as 
to whether an amplifier can be built that will amplify a 1-mv signal, with 
a specified source resistance, to a sufficiently high power level to operate 
a loud speaker or a pen recorder. Certainly enough cascaded stages can 
be used to provide any theoretical power gain. Excellent shielding and 
elimination of undesired coupling may be required to obtain stability at 
the required gain, but this requirement can almost always be met. The 
answer to the problem hinges basically on whether an amplifier is available 
which has sufficiently low internally generated noise so that the noise will 
not mask or obscure the signal. 

Since the first stage of an amplifier amplifies its own noise as well as 
the input signal, the combined noise and signal power available as an 
input signal to the second stage is usually large in comparison with the 
noise power generated by the second stage. Therefore, the first stape is 
the crucial one in considering noise contribution in an amplifier. Inductive 
reasoning shows that if the stages of a cascade amplifier each generate 
equal noise power and have equal power gains, the contribution of a given 
stage to the total noise power in the output of the amplifier is N c — 
NJG { ^~ V , where N is the noise generated in the amplifier, G„ is the 
power gain of each amplifier stage, and // is the numbered position of the 
stage (Fig. 9.1). 



Cascaded Amplifiers 



373 



The purpose of this section is to discuss the sources of noise, associate 
these sources with tube and transistor amplifiers, and gain some insight 
into the design of low-noise amplifiers. A figure of merit for amplifier 
noise will be defined. Noise sources are usually classified in three cate- 
gories: Thermal noise, diode noise, and 1// noise. 

Thermal noise, sometimes called Johnson noise, results from the random 
motion of charge carriers in a conductor. These random carriers produce 
a small net current in one direction one instant and in another direction the 
next instant. If these thermal currents are resolved into frequency com- 
ponents, they represent a continuous spectrum of essentially uniform 

R 

Vv\ -° 



@S„— y/AKTRAf -^JAKTGAf () = 



(a) W 

Fig. 9.21. Equivalent circuits for a thermal noise source. (a)The"venin's; (b) Norton's. 

amplitude from very low frequencies through the microwave range. There- 
fore, if a thermal noise signal were amplified by a noiseless amplifier, 
the noise power in the output of the amplifier would be proportional 
to the bandwidth of the amplifier. 

Nyquist 7 postulated that the maximum noise power which can be 
transferred from a normal (noisy) resistor to a noiseless resistor is 



p = kTkf 



.(9.77) 



where k is Boltzmann's constant and A/ is the bandwidth in cycles per- 
second. This relationship has been verified experimentally and is not 
surprising since kT is the energy equivalent of absolute temperature T. 
Since maximum energy transfer occurs when the load resistance is equal 
to the source resistance, the noise equivalent circuits of Fig. 9.21 can be 
drawn. A simple calculation will show that either circuit will transfer 
the maximum power (under matched conditions) given by Eq. 9.77. Of 
course, there are no such things as noiseless resistors (unfortunately) and 
when two resistors of equal temperature are connected together there is 
no net transfer of energy. The equivalent noise source voltage of a 
7 H. Nyquist, "Thermal Agitation of Electric Charge in Conductors," Phys. Rev., 
Vol. 32, p. 110,1928. 



3'"* Electronic Engineering 

resistor is, therefore 

v n = y/4kTR A/ (9.78) 

where v n is the rms value of the sum of all the noise voltage components 
in the frequency band A/. 

A diode (either thermionic or semiconductor) is a noise source because 
each emitted electron or injected carrier is a random event and small 
random fluctuations occur in the diode current which we have heretofore 
considered constant if the bias is fixed. These current fluctuations, like 
thermal noise, have a uniform and continuous frequency spectrum and 
therefore both diode and thermal noise are known as white noise. (In 
the visible spectrum, a continuous frequency spectrum produces white 
light.) Schottky postulated that the rms diode noise current in a given 
frequency band A/can be represented by a current generator of magnitude 



l n = sllql^f (9.79) 

where I is the average or Q-point value of the diode current. Equation 
9.79 does not hold for space-charge limited thermionic diodes because the 
space charge tends to smooth out or alter the randomness with which the 
electrons were emitted from the cathode. 

The third type of noise is known as 1// noise because its magnitude is 
approximately inversely proportional to frequency. This type of noise 
is effective only at lower audio frequencies, hundreds of Hertz or 
less. This 1// noise is caused by a phenomenon known as cathode 
flicker in vacuum tubes, and results primarily from surface leakage in 
transistors and semiconductor diodes. This type of noise varies quite 
widely among units of the same type device and has not been theoretically 
characterized. Recent improvements in transistor surface treatment have 
greatly reduced the 1// noise in comparison with earlier models. In 
applications which require a low noise amplifier the transistor or tube 
should be hand picked for low 1// noise. 

A noise figure of merit for an amplifier is the spot noise figure F, which 
is the ratio of the noise power delivered by an amplifier (over a narrow 
frequency band) to a load to the noise power that would be delivered if 
the only noisy component were the source resistance R s at T = 290°K 
(i.e., 290°K is defined as the standard reference temperature). Thus an amp- 
lifier which contributes no noise to the signal being amplified has a noise 
figure F ■= 1 (or F = db). The amplifier always contributes some noise; 
therefore, .Fis always greater than 1, or greater than db. 

The noise contributed by an amplifier can be represented by an equiv- 
alent rms noise voltage v n at the input of the amplifier, as shown in Fig. 
9.22. The equivalent noise voltage v n includes the noise generated by the 



Cascaded Amplifiers 



375 



source resistance R s . Therefore, the resistance R s in Fig. 9.22 is noiseless. 
The rms noise voltage v n can be characterized as the noise voltage pro- 
duced by a resistance R n at 290°K. Then the noise figure of the amplifier 
is RJR S , as is shown by the following relationships. First, the rms input 
noise current is 



In = 



V4fcTi? w A/ 
R s + Ri 



(9.80) 



R s + Ri 

where R { is the input resistance of the transistor. The rms noise output 
voltage is 



= iJjiRr = 



yj4kTR n A f 
R s + /?, 



G,K, 



(9.81) 






O 



) u s (Signal) 



Noiseless 
amplifier 



'Rr. 



Fig. 9.22. The representation of amplifier noise by an equivalent noise source. 

The rms noise voltage in the output which results only from the resistance 
of the source is 






J4kTR s M 
Rs + Ri 



GiR! 



(9.82) 



The noise power in the load is v n ?\R L , whereas if the amplifier were 
noiseless the noise power in the load would be (v nos ) 2 IRL- By definition, 
the spot noise figure is the ratio of these two noise powers. Then (using 
Eqs. 9.82 and 9.81, the spot noise figure is 

F _ v n» _ Rn (9.83) 

(tnosf Rs 

Observe from Eq. 9.83 that the noise figure F is independent of the load 
resistance R L . Also, the input resistance R t does not appear explicitly 
in the equation. However, the equivalent noise resistance R n is a function 
of the transistor input resistance. 

The noise sources of a transistor will be considered next, then a re- 
lationship for the equivalent noise resistance R n will be obtained. There 
are two major sources of noise in a transistor — the diode noise of the 
emitter current which is injected into the base region and the effective 
ohmic base resistance r b . The emitter diode current divides into the collector 



376 



Electronic Engineering 



current I c and the base current I B as discussed in Chapter 4. Thus three 
noise generators are shown in the hybrid-7r equivalent circuit of Fig. 9.23 
to represent these major noise sources. The equivalent noise voltage of 
the driving source is also shown. Minor noise sources which are not 
shown include the ohmic resistance of the emitter and collector regions 
and the saturation (diode) currents I co and I E0 which flow across the 
collector and emitter junctions, respectively. These sources are all 
negligible, providing that the g-point currents are large in comparison 
with the thermal saturation currents, which may not be the situation in a 
germanium transistor at moderate or high temperatures. 



<s> 



«ns 



"nb 



n 



-© — JW- 



InbQ) 



~Ci v 



gmV 



Q) O 



Fig. 9.23. The hybrid-w equivalent circuit of a transistor showing the major noise 
sources. 

Observe from Fig. 9.23 that the source noise voltage v ns , as well as the 
transistor noise sources v nb and l nb , must be amplified by the transistor 
before their noise contributions can be compared with the collector 
current noise component i nc . Also, Eq. 9.78 showed that a noise voltage 
is proportional to the square root of a resistance and Eq. 9.79 showed that 
a noise current is proportional to the square root of a quiescent diode 
current. Therefore, the following conclusions can be drawn: 

(a) The contribution of the collector current noise component /„„ is 
small if I c is small and the transistor current gain is high. 

(b) The contribution of the base current noise component l nb is small 
if the transistor current gain is high (thus permitting small I B ) and 
I EO is small. 

(c) The contribution of the base resistance noise component is small 
if r b is small in comparison with the source resistance R s . 

From these conclusions, it seems clear that a silicon transistor with high 
h FE at a low value of I c would have a low noise figure when driven from 
a high-resistance source (R s » r b ). 



377 
Cascaded Amplifiers 

In order to determine the noise figure for a specific transistor operated 
at a given Q point and driven by a specified source, a relationship will be 
developed for the equivalent noise resistance R n . First, an equivalent 
noise voltage v n in series with R s (see Fig. 9.22) will be determined which 
will produce the same effect at the output as all the noise generators of 
Fig 9 23 combined. In combining noise sources, it should be noted that 
independent sources are uncorrelated (that is, there is no fixed relationship 
between the phases of their relative frequency components) and, therefore, 
the total rms voltage is the square root of the sum of the squares of the 
individual rms contributions. Then 

v n * = v n * + v nb *+(v' nb y + (v' nc )* (9.84) 

The voltage source (v' nb ) results from the current source i nb of Fig. 9.23. 
This current source can be transformed to an equivalent voltage source to 



n 



-&- 



v Cj^ mU 



Fig. 9.24. An equivalent circuit used to determine v'„ c in terms of ! nc . 

the left of the imaginary terminals b'E in Fig. 9.23 by obtaining the open 
circuit voltage at the terminals b'E with the other voltage sources turned 

off. Then, 

v nb = Ur b + R s ) (9-85) 

The voltage source (v' nc ) is the voltage required to produce /„,. in the 
output. Figure 9.24 is drawn as an aid in determining this equivalent 
voltage. From this circuit is it seen that 



r _ v'ncZygr, 



(9.86) 



V „ r = 



(9.87) 



R a + r b + Z„ 
where Z, is the parallel combination of (h fe + IK and l/ywCj. Then 

r. f (R , + r b + Z„) 
Z„g m 
Substituting Eqs. 9.87 and 9.85 into Eq. 9.84, v n 2 becomes 

v^ = vJ + v nb 2 + ' nb (r b + Rs) + j— ^ ( 9 - 88 ) 



378 



Electronic Engineering 



The equivalent resistances and diode g-point currents can be substituted 
for the voltage and current sources. Then Eq. 9.88 becomes 

4kTR n A/ = 4kTR s A/ + 4fcl> 6 A/ 

+ 2qI B ^f{r i + R a f + 2^cA/(* + r 6 + Z T ) 2 



and 



R n = R s + r b + -f- I B (r h + tf s ) 2 + 



2kT 



Then, 



F = 1 + -*- + -S- J B 
K, 2feT 



(r» + Rsf 



+ 



j (R. + r b + Z,) 2 
2kT ° (g m Z r f 

q . (R s + r 6 + Z,) 2 



(9.90) 



(9.91) 



R. 2kT R s (g m Z„y 

The right-hand term of Eq. 9.91 is a function of frequency because Z w , 
which includes IjjmC^ is a function of frequency. As the frequency 
approaches f f and Z r decreases significantly, the noise factor begins to 




Fig. 9.25. Transistor noise as a function of frequency. 

increase, since Z T is in the denominator. However, since Z ff also appears 
in the numerator as a sum with R s and r b , the noise factor will not increase 
appreciably until Z n becomes smaller than R s + r b , which may be at a 
frequency considerably above f p , depending on R s . A sketch of noise 
figure as a function of frequency for a typical transistor is given in Fig. 
9.25. The 1// noise accounts for the rise at low frequencies. Equation 
9.91 shows that although the noise contribution of r b is decreased as R s 
is increased, the contributions from both the base and collector com- 
ponents of emitter current increase as R s is increased. Therefore, there is 
an optimum value of R s for a given set of transistor parameters. This 
optimum source resistance is the same order of magnitude as the input 
resistance of the transistor. Note also that the last term of Eq. 9.91 has 
the product g m Z f in its denominator. At frequencies below /„, Z T is 
approximately equal to (h fe + \)r e , and r e is approximately \jg m . There- 
fore, an improvement in noise factor will occur as I c is reduced until a 
point is reached where (h fe + l) 2 decreases as rapidly as I c . 



Cascaded Amplifiers 379 

Transistor manufacturers frequently supply constant noise figure 
contours for their low-noise transistor types. These contours give 
optimum values of quiescent current as a function of driving source 
resistance (Fig. 9.26). 

20 K 



10 K 



5K 



2K 



2 IK 



X 500 



200 



100 



















\5 


\X 


njH^ 


\17 


/=lKHz N 
A/ =200 Hz 
V C£ =10v 


\3 






















3\. 


Noise 
figure 
(db) 










\5 












\7 \ 






; 





10 20 50 100 200 500 1000 

I c -Collector current-Aia 

Fig. 9.26. Constant noise-figure contours for a typical low-noise transistor (2N2443). 



PROB. 9.20. A given silicon transistor has r h = 200 CI, h u = 150, h FE = 100 
at the Q point V CE = 5 volts, / c = lma and the frequency 1 KHz. Determine 
the 1 KHz spot noise figure at this Q point if the driving source resistance is 
2 Kft. Assume that the temperature is 300°K. The beta cutoff frequency is 
100 kc. Answer: F = 1.6. 

PROB. 9.21. The transistor of Prob. 9.20 has the Q-point collector current 
reduced to 0.1 ma, at which h fe = 90 and h FE = 60. Assuming all other 
parameters remain unchanged, determine the 1 KHz noise figure at this Q point. 
PROB. 9.22. Determine the spot noise figure of the transistor of Prob. 9.20 at 
the I c = 1 ma Q point at/ = 500 KHz. 

The field-effect transistor has only the channel resistance and surface 
leakage as principal sources of noise, since the only diode current is the 
thermal saturation current of the reverse biased gate diode. Therefore, 
the noise figure of a silicon FET may be very low. Since the diode current 
is very small, the medium frequency (1 KHz) noise figure is determined 
primarily by the ratio rjR s where r ch is the channel resistance. A typical 
value for F is 1.1 for high values of source resistance (of the order of 1 
megQ). The 1// noise is present, however, and the noise figure rises at low 



380 Electronic Engineering 

frequencies. Also, there is some diode current, so the noise figure increases 
as the power gain decreases at high frequencies. 

The triode vacuum tube has primarily diode noise which results from 
the random electron emission from the cathode. However, as mentioned 
in connection with the diode, the space charge surrounding the cathode 
tends to suppress the randomness of the emission current and reduces the 
tube noise well below the value which would occur in an emission limited 
tube. This smoothing effect complicates the derivation of a theoretical 
noise figure determination. However, an expression for the equivalent 
grid circuit noise resistance has been developed 8 for the triode and 
experimentally verified. This expression is 

R n = — (9.92) 

6m 

This noise resistance does not include the source resistance. Therefore, 
the noise figure F would be determined from the ratio (R s + R n )/R s , and 
high values of source resistance give low-noise figures. 

Like the transistor, the vacuum tube has a 1// noise known as cathode 
flicker which increases the noise figure at low frequencies. Also, the noise 
figure increases as the frequency becomes high and the electron transit 
time reduces the gain of the amplifier. The additional high frequency 
noise is known as induced grid noise. Low-noise triode tubes and low- 
noise transistors have noise figures of approximately equal magnitude. 

The tetrode or pentode tube has a source of noise in addition to those 
found in triodes. This noise, known as partition noise, results from the 
random division of the space current between the screen grid and the 
plate. The noise of a tetrode or pentode can be expressed as an equivalent 
grid circuit resistance by the following equation 

R n ^ + f^ (9.93) 

am ' Kom 

where I K is the average cathode current. The second term on the right- 
hand side of Eq. 9.93 accounts for the partition noise. 

The noise generated by a vacuum tube may be much higher than the 
values obtained by the foregoing equations because of defects in an 
individual tube or circuit. For example, gas in a tube causes random 
collisions which increase the noise. Also, faulty electrode contacts may 
generate noise. Heater-cathode interaction may cause noise and hum 
unless the bias between the heater and the cathode prevents an exchange 

8 K. R. Spangenberg, Vacuum Tubes, McGraw-Hill, New York, 1948, pp. 310-312. 



Cascaded Amplifiers 



381 



of charge carriers between the two. Usually low-noise tubes are hand 
picked after noise tests have been made. 

A noise test may be easily conducted on an amplifier if a noise diode 
is used as a signal source. The noise diode is connected in parallel with 
the desired source resistance as shown in Fig. 9.27a. The noise diode 
must have a high internal impedance compared with R s in order to act 
as a true current source. Therefore, a temperature limited vacuum 
diode is usually used. First the noise output voltage V no is measured 
with an rms activated meter while the noise diode is turned off. Then 
the output noise power results only from the equivalent amplifier noise 

Bs 
yW o 



© 



Amplifier 



VnD 



-yj4KTR n bf 
= -^j2qI D AfR s 



Fig. 9.27. A circuit used to determine the noise figure of an amplifier, (a) Block 
diagram; (b) equivalent input circuit. 

source v n (Fig. 9.27b). Next, the diode current is increased by adjusting 
the supply voltage V DD until the output voltage increases by a factor of 
*Jl, which indicates that the noise power in the output has doubled and 
the Thevenin's equivalent voltage of the diode noise source v nD is equal 
to equivalent noise voltage v n . Then, from Fig. 9.276, 



and 



2qI D HfR? = AkTR n \f 



K„ = 



2kT 



(9.94) 



(9.95) 



But the noise figure F is 



R, 2kT 



(9.96) 



At the standard reference temperature T = 295°K, qj2kT = 20, and 



F = 2W D R S 



(9.97) 



The noise figure obtained by the procedure described will give an 
integrated noise figure over the pass band of the amplifier or meter, 
whichever is less. A filter can be placed in series with the output meter 
to determine the spot noise figure in the pass band of the filter. 



-* 82 Electronic Engineering 

PROB. 9.23. A given dynamic microphone has the following specifications: 
R = 10 KO; output voltage = — 56 db below 1 v open circuit; frequency 
response =40 to 15,000 Hz. Design an amplifier, using 2N2712 transistors, 
that will provide l.Ovrms across a 10 KH load. Provide a pass band of 
approximately 20 to 20,000 Hz in the amplifier. Determine the approximate 
signal to noise ratio for your amplifier and express the noise as a number of 
db below the signal. Neglect 1// noise. 

PROB. 9.24. Repeat Example 9.2 for transistors with the following charac- 
teristics : 

h ie = 2000 n 

h re =j5 x 10- 3 

h fe = 50 

h oe = 10 +/20 

PROB. 9.25. Repeat Example 9.2 (using the transistors listed in the example) 
but replace the inductively coupled circuits between transistors T x and T 2 and 
between T 2 and T a with tuned circuits which contain tapped coils. 



io 



Large-Signal Amplifiers 



A typical amplifier consists of several stages of amplification. Most of 
these stages are small-signal, low-power devices. For these stages efficiency 
is usually not of major importance, distortion is negligible, and the 
equivalent circuits accurately predict their behavior (Chapters 7 to 9). In 
contrast, the final stage of an amplifier (and in some cases additional 
driver stages) is usually required to furnish appreciable signal power to its 
load. Typical loads include loudspeakers, antennas, positioning devices, 
and so on. These amplifiers are commonly called power amplifiers. 
Because of this relatively high power level, the efficiency of the power 
amplifier is important. Also, distortion becomes a problem because the 
amplifier parameters vary appreciably over the signal cycle. For this 
reason, the equivalent circuits are only rough approximations and graphical 
methods assume increased importance. 

10.1 CLASSIFICATION OF LARGE-SIGNAL AMPLIFIERS 

Many of the large-signal amplifiers are driven so hard by the input signal 
that amplifier current is either cut off or in the saturation region during a 
large portion of the input cycle. Consequently, a system for designating 
various operating conditions has been evolved. This designating system is 

383 



384 



Electronic Engineering 



Quiescent 
input voltage 





<c) 
Fig. 10.1. Classification of large-signal amplifiers. 
B amplifier; (c) class C amplifier. 



Quiescent 
input voltage 



(a) Class A amplifier; (b) class 



illustrated in Fig. 10.1. In this diagram, the input signal is assumed to have 
a sinusoidal waveform. The dynamic transfer characteristics are used to 
relate the input voltage and output current. Thus, if the output current 
flows for 360° of the input cycle as shown in Fig. 10.1a, the device is 
designated as a class A amplifier. Similarly, if output current flows for 
more than 180° but less than 360° of the input cycle, the device is called a 
class AB amplifier (not shown). When the output current flows for ap- 
proximately 180° of the input cycle as shown in Fig. 10.16, the device is 
biased at cutoff and is designated as a class B amplifier. Finally, if the 
output current flows for less than 180° of the input cycle as illustrated 
in Fig. 10.1c, the device is referred to as a class C amplifier. 1 In addition to 

1 The class C amplifier is normally used only in tuned amplifiers and therefore will 
not be considered in this chapter. However, the chapter on large-signal tuned amplifiers 
will discuss class C amplifiers. 



Large-Signal Amplifiers 



385 



the foregoing designations, the subscript 1 is given to a vacuum tube circuit 
if the control grid is never driven positive. If the control grid is driven 
positive, a subscript 2 is given to the designation. Obviously, the sub- 
scripts have no meaning when applied to transistors, since the transistor 
base always draws current unless the transistor is cut off. 

From the foregoing designations, the amplifiers that have been con- 
sidered in the small-signal devices are class A x amplifiers for vacuum 
tubes and class A amplifiers for transistors. These amplifiers are usually 
referred to as class A with the subscript understood. Similarly, a vacuum 
tube that is biased at cutoff but that has a large enough input signal to 
drive the control grid positive is referred to as a class B 2 amplifier. 
For class B amplifiers with no subscripts, the amplifier is assumed to be 
class B 2 . 

PROB. 10.1. A 6J5 tube is connected through a resistive load with R L = 20,000 
H to a 300-v Vpp supply. What class of operation does this tube exhibit if (a) 
v T = —6 + 5 sin cot; (b) v t = —6 + 7 sin cot; (c) vj = —18 + 16 sin cot; 
(d)v r = -30 + 34 sin tot. 

PROB. 10.2. Sketch the output waveforms for the input signals listed in Prob. 
10.1. The characteristic curves of a 6J5 tube with positive as well as negative 
control grid voltage is given in Fig. 10.2. 




100 



200 300 

Plate voltage in volts 

Fig. 10.2. 6J5 characteristic curves. 



400 



500 



386 Electronic Engineering 

10.2 DISTORTION AND POWER OUTPUT CONSIDERATIONS 

As already noted, the equivalent circuits provide accurate solutions to 
transistor and vacuum tube problems when the signal magnitudes are 
small. Unfortunately, as the magnitude of the signal increases, the accuracy 
of the equivalent circuits decreases. This loss of accuracy is due to the 
fact that the circuit parameters (r p , g m , h ie , h fe , etc.) are not linear. (This 
nonlinear behavior has been emphasized by Figs. 5.22 and 5.26.) As a 
result of these nonlinear elements, graphical solutions must be used if 

accurate results are desired and a large 
signal is present. 

In order to gain a clearer insight into 

the effects of nonlinear parameters, we 

will consider the collector current of a 

h f' ^^ transistor which has a nonlinear h u . 

Curves depicting the variation of the 

h parameters with emitter current or 

collector voltage (similar to thos'e in 

i B — *- Fig. 5.22 or Fig. 3.2) are often given by 

Fig. 10.3. A typical curve of h„ the transistor manufacturer. Since i c 

vs i B . depends on base current, a curve of 

hfe vs >b (F'g- 10.3) can be determined. 
From the ideas developed in analytical geometry, an expression can be 
found that will describe the curve given in Fig. 10.3. In general, this ex- 
pression will have the form 

hfe = a + a i's + C2'b 2 + <W H (10.1) 

where a , o,, a 2 > etc. are constants. The current which flows in the collector 
circuit is proportional to h u i B . Thus, the collector current can be written 
as 

'c = Bhfe'B = A i B + AJ^ + A 2 i B s + A 3 i B * + ■■■ (10.2) 

where B is the proportionality constant and A = a B, A t = a t B, etc. 
If i B contains a d-c and a sinusoidal component 

>b = h + h sin cot (10.3) 

When this value of i B is substituted into Eq. 10.2, the first term on the 
right side of Eq. 10.2 becomes 

A o'B — A <s I B + A o [ b sin l0t ( 10 - 4 ) 

The second term becomes 

a i>b 2 = ^i(V + 2I B I b sin cot + V sin 2 cot) (10.5) 



Large-Signal Amplifiers 387 

However, a trigonometry identity states 

. , (1 — cos 2x) , 1rt ,. 

sm 2 x = i (10.6) 

2 

Thus, Eq. 10.5 can be written as 

(72 r 2 \ 

V + — + 2I B I b sin cot - -*- cos 2a>t I (10.7) 

From Eq. 10.4, the ^ /jj term produces a d-c component and a sinusoidal 
component in the collector current. From Eq. 10.7, the A^ 2 term 
produces a d-c component, a sinusoidal component, and a second harmonic 
component in the collector current. The A 2 i B 3 term will produce d-c, 
sinusoidal, second-harmonic, and third-harmonic terms. It follows that 
the higher terms will produce higher harmonics. The production of these 
harmonic terms is referred to as distortion. The effect of second harmonic 
distortion on the output waveform is shown in Fig. 10.4. Notice how the 
output signal has been modified or distorted from the original sine wave. 

If h u (or g m of a tube) were the only nonlinear parameter, it would be 
possible to use an equivalent circuit with a nonlinear h u (or g m ) and still 
to obtain results without too much complication. However, since not 
only h u and g m but also h ie , h oe , h re , fi, and r p are nonlinear, the equiv- 
alent circuits become too complex to be useful. Thus, it is usually much 
easier to use a graphical solution for nonlinear operation. 

It is interesting to note that some of the nonlinearities in tubes and 
transistors are of compensating nature. Thus, for a vacuum tube amplifier 
(Fig. 5.26), g m decreases as the control grid becomes more negative. 
However, note that r p increases as the control grid becomes more negative. 
Thus, the shunting effect of r p decreases and a larger portion of the 
current from g m v g goes through the load resistor. These effects do not 
completely cancel, but at least the gain does not decrease as much as 
the decreasing g m may indicate. Similar compensating effects will also 
be noted in transistor amplifiers in the next section. 

PROB. 10.3. Find an expression similar to Eq. 10.7 for the term A^. Assume 
i B = I B + It sin cot. 

As indicated, the output current waveform may contain a number of 
harmonics in addition to the fundamental component. Also, if the load 
is a pure resistance, the output voltage has the same waveform and hence 
the same harmonic content as the output current. Thus the output 
current (or voltage) may be expressed as a Fourier series. This series will 



388 



Electronic Engineering 



*o 



l O l Q 



Fundamental 




Second harmonic 
f d.c, or average 



Time 



(a) 




Time >■ 

(b) 

Fig. 10.4. The effect of second harmonic distortion, (a) Frequency components; (b) 
Total current. 

have only cosine terms if the time reference (/ = 0) is chosen at a point of 
symmetry as shown in Fig. 10.4. Then 



i Q = Iq + h cos tot + I 2 cos 2cot + I 3 cos 3cot + • 



(10.8) 



The values of I , I lf I 2 , etc. can be determined from a graphical plot of 
the output waveform. In fact, some of the harmonics may be deduced 



Large-Signal Amplifiers 389 

to be negligible by an examination of the output waveform. For example, 
waveforms with the general shape shown in Fig. 10.4 can be con sidered 
to have predominantly second harmonic distortion because successive 
half cycles are dissimilar. Thus waveforms of this general shape can be 
approximately represented by the equation 

i = I + I Y cos tot + I 2 cos 2cot (10.9) 

where 7 X represents the peak value of the fundamental component and I 2 
represents the peak value of the second harmonic component. Now there 
are three unknown quantities /„, I lt and 7 2 which must be determined, so 
three independent equations involving these three unknowns must be 
found. For example, at time r = 0, (Fig. 10.4) i = z' max and Eq. 10.9 
becomes 

'max = h + h + h (10-10) 

Similarly, at time / = njlco, i Q = i Q and Eq. 10.9 becomes 

i Q = h~h (10.11) 

Also, at time t = it/cd, i = / min and Eq. 10.9 becomes 

'min = h ~ h + h (10.12) 

These three equations can now be solved simultaneously to yield the 
values of 7 , I lt and I 2 in terms of the graphically determined values of 

'max- 'Q. an d 'min- 

T 'max — 'min s.~ « ^x 

(10.14) 

(10.15) 

The percentage of second harmonic distortion in the amplifier is 

%sec = ^ x 100 (10.16) 

PROB. 10.4. Obtain Eqs. 10.13, 10.14, and 10.15 from Eqs. 10.10, 10.11, and 

10.12. 

PROB. 10.5. One section of a 12AT7 triode is connected as a common emitter 

amplifier with a load resistance R L of 20 Kfl. The V PP supply is +300 volts. 

If Vj = — 2 + 2 sin cot, find the percentage of second harmonic distortion in 

this amplifier. Answer: 9.45% second harmonic. 



'1 






2 








h 


= 


'max 


+ 


'min 


-2i 









4 






/„ 


= 


'max 


+ 


'min 


+ 2/ 


Q 






4 







390 



Electronic Engineering 



A second type of distortion which commonly occurs in push-pull amplifiers 
(which are discussed later in this chapter) is shown in Fig. 10.5. In this 
type, successive half cycles are similar, so even harmonics are negligible, 
but the peaks of the wave are flattened, so the third harmonic may be 
assumed to be the principal distortion component as shown. Then, with 
the time reference selected as shown in Fig. 10.5, only sine terms appear 
and the output current may be represented approximately by the series 



'o = A) + A sm w ' + h sm 3<uf 



(10.17) 



'max — 




Time 
Fig. 10.5. Typical third harmonic distortion. 



Again, there are three unknowns, so three equations must be written. At 
time t = 0, i = i Q and Eq. 10.17 becomes 



Iq = h 



At t = 77/600, i = i n/e (See Fig. 10.5) and Eq. 10.17 becomes 



l ir/6 ~~ ^0 + ~ + -^3 



(10.18) 



(10.19) 



where / t/6 is the output current that flows when the input quantity is one- 
half its maximum value, which occurs at 77/6 radians since sin 77/6 = J. 
At t = 7TJ2(o, i = / max and Eq. 10.17 becomes 



=h+h~h 



(10.20) 



Equation 10.18 determines the value of I . When this value is substituted 
into Eqs. 10.19 and 10.20, these equations can be solved simultaneously 



Large-Signal Amplifiers 391 

for I x and I 3 . Thus 

1_ 3 

and 

_ ^'V/6 ~ 'max — Jq ^q 22) 

3 3 

PROB. 10.6. Derive Eqs. 10.21 and 10.22 from Eqs. 10.18, 10.19, and 10.20. 

The foregoing method of analysis can be used to obtain any given 
number of harmonics in the output of an amplifier. Of course, the number 
of independent equations found must be equal to the number of significant 
frequency components in the output current (or voltage) waveform. 
Equations which include all components through the fourth harmonic 
may be found in the literature. 2 

The ideal large-signal amplifier should have high-power output and 
low signal distortion. Unfortunately, these two characteristics are not 
compatible. Hence, a compromise must be reached between maximum 
power output and minimum distortion. Typical plots of power output and 
distortion vs load resistance R L for vacuum tube amplifiers are shown in 
Fig. 10.6. (As will be found in Section 10.3, the distortion of a transistor 
amplifier depends on the impedance of the current source driving the 
transistor as well as on the load resistance.) In both cases in Fig. 10.6, the 
distortion is quite high for maximum power output. However, the dis- 
tortion can be reduced to acceptable levels by a modest reduction of the 
power output. 

Figure 10.7a and Fig. 10.76 show why the power output and the dis- 
tortion of vacuum tube amplifiers vary with the load resistance in the 
manner given in Fig. 10.6. The assumption is made that the grid voltage 
varies between zero volts and the cutoff voltage. Also, the plate voltage 
V PP is assumed fixed. Since the signal power output is proportional to the 
product of the plate voltage variation and the plate current variation, the 
power delivered to the load is proportional to the area of the right tri- 
angle which has the load line as its hypotenuse. Thus, for the smallest load 
resistance shown in Fig. 10.7, the power output is represented by the 
area of the triangle V PP — A — A'. Notice that for the tube amplifiers, 
the power output increases as the load resistance is increased, and the load 
line is shifted from the position V PP - A to the position V PP — B. A 
further increase of load resistance causes the load line to shift toward the 
position V PP — C and the power output decreases accordingly. 

2 F. E. Terman, Radio Engineers Handbook, McGraw-Hill, New York, 1943. 



392 



Electronic Engineering 







\ 
















\ 
















y- 
















/\ 
















/ \ 
















i 


\ 






<°v 




• 3 






\ 


























3 






\ 
















^4i 








0> 

5 






\%.. 








£ 






\ 


f* 
\ 

\ 
















*- 
















*«» 


































""~—-— ■ 


- 



Load resistance-ohms ■ 
(a) 













■~5°!!«r 




3* 


^s 


V / 




vvO^ 


harmonic 


— 


3 

o 












2 J^1 


jvoo^— 



















Load resistance-ohms *■ 

<b) 

Fig. 10.6. Power output and distortion vs R L for some electronic devices, (a) Triode 
tube; (b) beam power tube. 



For the triode tube the plate current swing continually decreases as the 
load resistance increases. Therefore, the variation of the tube parameters 
and hence the distortion continually decreases as the load resistance 
increases. On the other hand, the plate current swing of the pentode does 
not decrease appreciably until the load line passes through the knee of the 
v a = curve. However, the second harmonic distortion does decrease as 
the knee is approached because the spacing of the plate characteristic 
curves becomes more uniform. The second harmonic actually approaches 
zero as the load line approaches the knee of the v a = curve. The 



Large-Signal Amplifiers 393 

dynamic transfer characteristic would appear as a slightly j-shaped, 
almost symmetrical curve for this value of load resistance. As the load 
resistance of the pentode is further increased, the distortion increases 
while the power output decreases, because the output wave is severely 
flattened on the positive half-cycle of the grid voltage swing. 

An upper current limit for the transistor amplifier is often specified, but 
this limit is usually so high that the collector dissipation determines the 



+20 



+ 10 




Small flr 



Medium R L 




B' A' V PP 
Plate voltage 
(a) 



C'B' 



A- V PP 
Plate voltage 
(b) 




Collector voltage 
(c) 
Fig. 10.7. The effect of R L on distortion and power output, (a) Triode tube; (6) pentode 
tube; (c) transistor. 

minimum value of load resistance that can be used. Therefore, the power 
output of the transistor amplifier increases as the load resistance is 
decreased, the limit being set by the dissipation capabilities of the tran- 
sistor. This increasing power output with decreasing load resistance 
appears to be contrary to the maximum power transfer theorem. However, 
this theorem is not being violated because the power of the driving source 
must be increased to obtain the higher power output. The power gain, 
which is maximum under matched conditions, actually decreases as the 
load resistance is decreased. Methods of analyzing distortion in transistor 
amplifiers will be given in Section 10.4. 



394 



Electronic Engineering 



When power output as a function of load resistance is considered, the 
tube and transistor cases differ primarily because of the maximum positive 
grid voltage limit imposed in the vacuum tube case. If the grid could be 
driven positive as far as desired, the upper plate current limit would be 
determined only by the plate dissipation capability of the tube. This 
statement may be verified by the observation of Fig. 10.7a and Fig. 10.76. 

In all amplifiers, care should be exercised to confine the operation 
within the limits of maximum dissipation ratings as well as within the 
maximum current and maximum voltage ratings. 



10.3 GRAPHICAL SOLUTION OF TRANSISTOR CIRCUITS FOR 
LARGE SIGNALS 

As indicated in Section 10.2, the input characteristics of a transistor as 
well as the transfer characteristics are nonlinear. Consequently, when 
transistor amplifier distortion is considered, the nonlinearity of the input 
and transfer characteristics must be considered simultaneously. Hence, 
the graphical method of solution previously used must be modified for 
large signals. This modified method will be illustrated by an example. 

Example 10.1. A transistor is connected as shown in Fig. 10.8. The input and 
collector characteristics of this transistor are given in Fig. 10.9. Plot the output 
current of this circuit if the input signal is as given in Fig. 10.8. 

The first step in the solution of this problem is to draw the load line on the 
collector characteristics of Fig. 10.9. Then the dynamic input characteristic is 
drawn as shown in Fig. 10.96. From the curves of Fig. 10.9, the dynamic 
characteristic curves can be drawn as shown in Fig. 10.10. The dynamic curve 
relating i c and i B (curve A) is derived from Fig. 10.9a, and the dynamic curve 
relating i B and v B (curve B) is taken from Fig. 10.96. 

The source voltage v z is shown in Fig. 10.10. This voltage must be reduced by 
the i B R drop through R getl to find the actual voltage at the base of the transistor. 



"7 




-0.1 
-0.2 
-0.3 
-0.4 
-0.5 
-0.6 



250 fl 




Fig. 10.8. A transistor amplifier. 



Large-Signal Amplifiers 

-200 



395 




-0.4 



-0.3 



-0.2 



-0.1 



-5 



-10 -15 

Collector volts 
(a) 



-20 



-25 







-3to-25v~^ 








Dyn 


"2.0 v-^" / 

„ c = -0.lv 
amic characteri 


r 7 

itic-^ 























-0.5 -1.0 -1.5 -2.0 

Base current (milliamperes) 

(b) 



-2.5 



Fig. 10.9. Transistor characteristic curves. 



396 



Electronic Engineering 



Accordingly, lines with the slope of R gejl are drawn from the vj value (point C 
for example) to curve B (point Q). The projection of this point (point C x ) to the 
i B axis (point C 2 ) yields the magnitude of i B for the given value of v z (point C). 
This value of i B (point C 2 ) is projected to curve A (point C3). The point of 
intersection on curve A (point C 3 ) is projected to the i c axis (point C 4 ) to give 



-150 



ic 



r^ 




Slope = Rgen 



Fig. 10.10. Large-signal graphical solution of a transistor circuit. 

the value of output current for the given value of voltage v T (point C). Other 
values of voltage vj are projected until the required current waveform can be 
drawn. 

If the desired output is voltage rather than current, curve A is plotted as 
v c vs i B instead of i c vs i B as in Example 10.1. The voltage gain of the 
stage can thus be found directly. 

Usually, curve B has more curvature than curve A. Therefore, the input 
characteristics are usually more nonlinear than the transfer characteristics. 
It is interesting to note that if curve A is the mirror image of curve B and if 
R gen is zero, the two nonlinearities will cancel and no distortion will be 
present in the circuit. Even though the two nonlinearities do not cancel, 



Large-Signal Amplifiers 397 

at least the curvature is such so as to minimize distortion. This compen- 
sating effect is, therefore, similar in action to the compensating effect of r v 
and g m in the triode tube which was mentioned in Section 10.2. 

It is interesting to note that the internal resistance (R gen in Fig. 10.8) of 
the signal source has an important effect on the distortion of a transistor 
amplifier. The effect of this source resistance is illustrated in Fig. 10.11. 




Fig. 10.11. The effect of signal source impedance on distortion in a transistor. 

In this figure, the solid lines show the solution for a large value of R gen . 
For large R gen , the collector current swing from minimum current to 
quiescent current is greater than the current swing from quiescent current 
to peak current. In contrast to this case, the dashed lines of Fig. 10.11 
show the solution if R gen is small. For small i? gen , the current swing from 
minimum current to quiescent current is smaller than the current swing 
from quiescent current to peak current. Obviously, some value of i? gen 
must exist such that the distortion is very small. In fact, the second 



398 



Electronic Engineering 



harmonic component may be zero. Thus, in case the driving source resist- 
ance is not optimum, the distortion of a transistor may be decreased by 
adding a resistor in series with the base (to increase the effective i? gen ) or 
in parallel with the base (to decrease the effective R gen ). True, these resistors 
decrease the power gain of the circuit, but this price must be paid for low 
distortion. 




Fig. 10.12. Arrangement for determining optimum R gen - 

The value of driving source resistance i? gen that will give (/ Cmax — i C(3 ) = 
Ocq — i'c min)> anc * therefore near-minimum distortion, can be determined 
for a given load line on a particular transistor. A relationship for this 
optimum value of 7? gen will be derived with the aid of Fig. 10.12. This 
figure is obtained by drawing the dynamic transfer and input character- 
istics for a given a-c load resistance as previously discussed. The values 
for i c max (at or near saturation) and i c min (at or near i B = 0) are 
established next. The value of i CQ is set at the mid-point between i c MAX 
and i c MIN . The corresponding values of i B max, h mix- *bq an d v BK MAX , 
v BEt,wx an d v BEQ are then graphically determined (Fig. 10.12). Also, 
the corresponding source potentials may be determined as follows : 

(10.23) 
(10.24) 
(10.25) 



V S MAX — V BE MAX + l B MAX°gen 
V SQ — V BEQ + 'Bergen 
V S MIN = V BE MIN ~t" l B MIN "gen 



Large-Signal Amplifiers 395> 

But v s is assumed to be a sinusoidal (or symmetrical) signal. Therefore 

V S MAX ~ "SQ = V 8Q - %IN (10.26) 

Substituting Eqs. 10.23, 10.24, and 10.25 into Eq. 10.26 

VbE MAX + v BEM\y + -Kgcnfe MAX + «B MIn) = ^( V BEQ + 'BO^gen) 

(10.27) 
Solving for i? gen , 

«gen — — . ~. »■ 

'j?MAX T 'BMIN Z, BQ 

If the calculated value of R gen is zero or negative, and therefore unrealiz- 
able, a minimum practical value of R gen should be used. 

PROB. 10.6. Sketch v c for the transistor in Example 10.1 by drawing curve A 

as v c vs i B . What other system could be used when i c is known? 

PROB. 10.7. If R L in Example 10.1 is changed from 250 CI to 1000 Ci, sketch i c . 

PROB. 10.8. Graphically find the per cent of second harmonic distortion in 

the output current of the transistor in Example 10.1. Assume the input voltage, 

Vj, is a sine wave with the same peak-to-peak value as the voltage v z used in 

Example 10.1. Answer: % sec = 35.2. 

PROB. 10.9. Repeat Prob. 10.8 if R sen is changed to 5000 ii. 

10.4 EFFICIENCY OF CLASS A AMPLIFIERS 

The plate or collector efficiency of a power amplifier can be found by 
dividing the output signal power by the input plate or collector source 
power. The average input power to the collector of a transistor P c in is 
equal to the collector supply voltage V cc times the average collector 
current I c or 

Pc m = Vcdc 00.29) 

Similarly, the average input plate power to a tube amplifier (P P ln ) is 

Pp m =V PP I P (10-30) 

In both cases of efficiency, the instantaneous signal power to the load is 
given by the instantaneous voltage times the instantaneous current. If the 
signal is a sine wave and the load is purely resistive, the signal power is 

Pont = Yl = I * RL (10.31) 

where V and I are the rms value of the output signal. Now, if Umax is 
the maximum voltage across the load and u MlN is the minimum voltage 



400 



Electronic Engineering 

across the load, the peak-to-peak signal voltage is » MAX - v ms . There- 
fore, if no distortion is present, 



V = 



"max ~ "min 



Hence, 



2(2) 



}A 



p out __ ("max — "min) _ ("max — "minX'max — »'min) 



(10.32) 



(10.33) 



SR L 8 

An example can help illustrate the procedure for calculating efficiency. 

Example 10.2. A circuit is connected as shown in Fig. 10.13. If the transistor is 
assumed to have no distortion, what is the maximum collector efficiency of a 
class A amplifier when a sinusoidal signal is applied ? 




Fig. 10.13. The circuit for Example 10.2. 

Since the collector circuit power input is constant (Eq. 10.29), maximum 
efficiency occurs when the collector voltage swing is maximum (Eq. 10.33). 
Since the quiescent collector voltage is V cc , the maximum collector voltage 
swing occurs when the collector is driven to zero as a minimum and 2 V cc as a 
maximum. (See Fig. 10.14.) Actually, the collector cannot be driven completely 
to zero but zero is the limiting value. 

For maximum efficiency, let i c be when v c is equal to 2 V cc . The maximum 
collector current will then be 



l C MAX 



2 V, 



cc 



rPR, 



(10.34) 



where n is the turns ratio of the transformer N^N^. The quiescent collector 
current is 



"cc 
n*Rr 



Therefore, the input power is 



^in — Vcch 



cc'c 



n 2 R r . 



(10.35) 



(10.36) 



Large-Signal Amplifiers 401 

The rms value of the output voltage across the transformer is 0.707 V cc , 

v 2 
cc (10.37) 



"out. — 



2n 2 R r 



Consequently, the maximum class A collector circuit efficiency is 



Vc = 



V cc *l2n*R L 



= 50% 



(10.38) 



P in Vcfl#R L 

As the foregoing example illustrates, the maximum efficiency for a class 
A amplifier is only 50%. This example used a transistor, but identical 




"o v cc 2V CC 

Collector voltage 

Fig. 10.14. Relations of voltage and current in Example 10.2. 

results are obtained for plate efficiencies of vacuum tubes. Since an actual 
amplifier cannot be driven from to 2 V cc , the efficiency of actual class A 
amplifiers is never as high as 50%. However, an examination of typical 
transistor characteristic curves will show that the transistor can be driven 
almost to v c = before current saturation occurs. Hence, transistor 
amplifiers can approach 50% efficiency. 

Pentodes have characteristic curves similar to the transistors but plate 
current saturation occurs at a relatively higher potential. Accordingly, 
the plate circuit efficiency of a pentode is usually lower than the efficiency 
of a transistor. When the input power to the screen and filament is 
considered, the overall efficiency of a pentode is considerably lower. 

The characteristic curves of a typical triode tube indicate that the plate 
voltage is usually quite high when the control grid is at potential. 
Accordingly, the plate circuit efficiency of a class A 1 triode is much lower 
than the plate circuit efficiency of a pentode. Again, when the filament 
power is also considered, the overall efficiency of a triode is quite low. 
Note that the plate circuit efficiency of a triode can be increased if the 
control grid is driven positive. 



402 Electronic Engineering 

The foregoing discussion assumed a sinusoidal input signal. If other 
input waveforms are assumed, the collector circuit efficiency can be much 
higher than the 50% indicated. In fact, the sinusoidal input signal has one 
of the lowest efficiencies possible. When the input signal is a square wave, 
the efficiency of a class A transistor amplifier may approach 100%. This 
very high efficiency is obtained when the transistor is driven to cutoff 
during one half-cycle and into saturation during the other half cycle. The 
collector dissipation is then very low during both half cycles because the 
product v CE i c is very low during both half cycles. This mode of operation 
is used in switching applications. In this mode of operation the collector 
dissipation increases greatly during the switching time. Therefore, the 
efficiency may be very high only in case the switching time is very short in 
comparison with the period. 

Strictly speaking, the excitation or driving power as well as the power 
loss in the bias source should be included when overall efficiency is con- 
sidered. However, the driving power is usually small in comparison with 
the collector or plate power input. On the other hand, the power loss in 
the bias resistors, especially emitter circuit resistors, may not be negligible 
and must be included when overall efficiency is considered. 

Collector or plate circuit efficiency is usually of much greater importance 
than overall amplifier efficiency because the plate or collector circuit 
efficiency determines the amount of power output which can be obtained 
from the amplifier. The relationship between the collector or plate 
efficiency and the power output will be derived in a following section. 

PROB. 10.10. Show that the maximum efficiency for a class A transistor 
amplifier with a resistive load in the plate circuit is 25 %. 

PROB. 10.11. Although a 6J5 is not primarily intended as a power amplifier 
device, calculate P- m and P out as well as overall efficiency if V PP = 300 v and 
P L = 10,000 H. Assume class ^ x operation and keep the output relatively linear. 
PROB. 10.12. The characteristics of a 6V6 are given in Appendix 3. Calculate 
P m and P out for a 6V6 if V PP = 300 v and R L = 15 £2. What is the maximum 
plate circuit efficiency? The load is transformer coupled with n = 20. The 
grid to cathode voltage is ( — 10 + 10 sin ait)v. Answer: P in = 17.4 w, P out = . 
4.4w,tj p = 25.2%. 

10.5 PUSH-PULL AMPLIFIERS 

The preceding paragraphs have indicated that the large-signal amplifier 
can produce considerable distortion of the excitation waveform. For 
example, the amplifier output signal may not be symmetrical even though 
the input signal is symmetrical, such as a sinusoid. This situation results 
because the changing parameters cause the amplifier to have different 
values of gain on alternate half cycles. A Fourier analysis shows that this 



403 
Large-Signal Amplifiers 

unsymmetrical waveform is rich in even harmonics. (Actually odd 
harmonics may also be present.) On the other hand, any periodic wave 
^hin hj is symmetrical about a neutral axi s contains no even harmonics. 
fhuTan amplifier which flattens the input wave equally on both half 
cycles produces odd harmonics but no even harmonics. 




Input SUllJS. 
signal 



(a) 



Input 
signal 




(b) 
Fig. 10.15. Typical push-pull amplifier circuits. 

A type of amplifier known as a push-pull amplifier can eliminate even 
harmonic distortion (generated in the amplifier) by providing a sym- 
metrical output signal. Of course, any second harmonic present in the 
input signal will be reproduced in the output. Some typical push-pull 
circuits are given in Fig. 10.15. In these circuits, the input signal current 
introduces a voltage into the secondary winding of the transformer T,. 
Since the center of this" secondary winding is grounded, one end of the 
secondary will become positive while the other end is going negative. 
Therefore, the input signal to the amplifier A 1 will be inverted in relation 
to the input signal to the amplifier A 2 . TJ jJS polarity inversion between 



404 



Electronic Engineering 



t he two exciting signals is necessary for proper operation of the push-pu ll 
amplifie r. These two input signals should also be equal in magnitud e. 
If interstage transformer coupling is undesirable, the transformer 7\ may 
be replaced by one of the phase inverter circuits of Section 10.8. 

The basic operating principles of the push-pull amplifier are illustrated 
by the equivalent circuits of Fig. 10.16. These circuits represent either 
transistor or vacuum tube push-pull amplifiers. The d-c circuit of Fig. 
10.16 is drawn for the quiescent condition; that is, no input signal. In 
this condition, the amplifier currents flow through the output transformer 



Amplifier 



A <p <E*o 




Amplifier 
A 2 U) 



d-c a-c 

Fig. 10.16. The d-c and a-c equivalent circuits for a push-pull amplifier. 

primary in opposite directions so that the mmf produced in the trans- 
former by these currents tend to cancel. When the amplifying devices are 
perfectly matched and the output transformer is precisely center-tapped, 
there is essentially no magnetic flux in the transformer under quiescent 
conditions. 

When a time-varying signal is applied to the input terminals of the 
push-pull amplifier, the signal components of amplifier current flow in the 
relative directions shown in the a-c equivalent circuit of Fig. 10.16. 
Observe that the signal current in the lower amplifier A 2 is reversed in 
comparison with the signal current in the upper amplifier A x . These 
relative current relationships, as mentioned before, occur because the 
excitation of amplifier A 2 is of opposite polarity to the excitation of 
amplifier A x . The a-c equivalent circuit shows that the signal current 
components produce mmf which add in the output transformer. There- 
fore, the total transformer mmf is the product of the signal current and the 
total number of primary turns. Hence, the power furnished to the load is 
the sum of the signal powers from the two tubes or transistors. Note that 
the signal current components do not pass through the power supply or 
the emitter biasing circuit. 



Large-Signal Amplifiers 



405 



The even harmonic distortion is cancelled in the push-pull amplifier 
output (assuming the circuit is balanced) because the negative half cycle 
of one tube or transistor adds to the positive half cycle of the other. 
Figure 10.17 illustrates the manner in which the even harmonics are 
cancelled. In this figure, the input signal 
is assumed to be sinusoidal. This can- 
cellation of even harmonic distortion 
can also be shown mathematically if 
the output current is represented as a 
Fourier series. The input signal voltage 
source is assumed to be v 1 = V sin cot. 
Then the output signal current of am- 
plifier A x is 

i\ = 5 X sin cot + B 2 sin (2cot + <j> 2 ) 
+ B a sin (icot + <f> 2 ) 
+ ---B n sin (ncot + h) (10.39) 
Since the input voltage of amplifier A 2 
is inverted (1 80° out of phase) when com- 
pared with that of A u t> 2 = V sin (cot+n). 
Then the output current of amplifier A 2 
is 

i 2 = B^ sin (cot + it) 

+ B 2 sin [2(cot + it) + <£ 2 ] 
+ B 3 sin [i(cot + n) + ct> 3 ] 
+ • ■ ■ B„ sin [n(cot + -it) 
+ *„! (10-40) 

Re-writing Eq. 10.40 and recognizing that an even multiple of n radians, 
when added to an angle, does not change the angle, we see that 

j 2 = B l sin (cot + n) + B 2 sin (2cot + <f> 2 ) + B a sin (Scot + tt + <&,) 

+ • • • B n sin (ncot + n-n + $„) ■ • • . (10.41) 

By comparing Eqs. 10.41 and 10.39, we can see that only the fundamental 
and odd harmonics have opposite polarities in the output currents and thus 
add in the output transformer. Conversely, the even harmonics are in 
phase and therefore cancel in the output transformer. 

The push-pull circuit can be analyzed by the use of the a-c equivalent 
circuit of Fig. 10.16. However, the two amplifying devices A x and A 2 
appear to be in series insofar as the signal components of output current 
are concerned. Therefore, the analysis is made easier by replacing the 




Fig. 10.17. Illustration of the can- 
cellation of even harmonic distortion 
in the output of a balanced push- 
pull amplifier. 



406 



Electronic Engineering 



current sources of Fig. 10.16 with voltage sources as shown in Fig. 10.18. 
In this circuit, N t is the number of turns in half of the primary and N 2 is the 
number of secondary turns. Then, assuming the circuit to be balanced, 
we see that 

k -"-». + (5 J K fR L (10 ' 42) 

The instantaneous power delivered to the primary of the power transformer 
is equal to i a times the reflected load resistance, or 

2v 



Pout = 



I2r + (27V 



'7m-J(th <-> 



)v = iR n 



)v = iR 



-r 




Fig. 10.18. A voltage source equivalent circuit for the push-pull amplifier. 



Equation 10.43 can be rewritten as 

4v 



Pout = 



2r + 4(NjN 2 fR L . 



'&)« 



(10.44) 



Dividing both the numerator and denominator of the bracketed term by 4, 
we have 



Pout = 



II 



R, 



(10.45) 



_(r /2) + (NJN^RlJ 

An equivalent circuit which will yield the power of Eq. 10.45 is given in 
Fig. 10.19a. The current source equivalent circuit is shown in Fig. 10.1%. 
These equivalent circuits represent a fictitious tube or transistor which has 
the same characteristics as a push-pull configuration. This fictitious tube 
is known as the composite tube. Similarly, the transistor represented by 
the equivalent circuit of Fig. 10.196 is known as a composite transistor. 

The composite collector or plate characteristics may be obtained from 
the respective individual characteristics as illustrated in Fig. 10.20. The 




N 2 >R L 2i 



<S> 



r„/2- 




(a) « 

Fig. 10.19. Equivalent circuits for the composite tube or transistor. 



100 



Plate volts 
200 300 400 500 600 70 




700 



600 



500 



400 300 

Plate volts 



200 



Fig. 10.20. Composite characteristics and load lines for push-pull tubes. 

407 



408 Electronic Engineering 

triode-connected 6L6, whose plate characteristics are given in Fig. 10.20, 
will be used to illustrate the method of obtaining the composite charac- 
teristics. One characteristic curve is inverted and the i P = axes are 
made to coincide. The two curves are adjusted until the quiescent plate 
voltages of the two curves coincide. In this case, the quiescent plate 
voltage is 300 v so the 300-v lines are made to coincide. Next, a d-c load 
line is drawn and the quiescent operating point is chosen for each indi- 
vidual tube. In this example, the d-c resistance of the transformer primary 
is assumed to (be negligible. Therefore the d-c load lines are drawn 
vertically and the quiescent operating point is chosen at the intersection of 
this d-c load line and the v G = —30 curve. 

Now, since the two plate currents subtract in the output transformer, 
the two currents can be subtracted graphically. For example, if no time- 
varying signal is present, both control grids will have — 30 v applied and 
the dashed curve for v GC = (Fig. 10.20) results. This curve is seen to be 
the difference between the two v a = —30 curves. Similarly, if the v G of 
tube 1 is —15 v, the v G of tube 2 is —45 v. The difference between these 
two curves is the v oc = +15 v composite curve. Similarly, the v oc = 
+30 v composite curve is the difference between the V a = 0-v curve of 
tube 1 and the V G = — 60-v curve of tube 2. Since the circuit is sym- 
metrical, the composite curves for v GC = — 15 v and — 30 v are mirror 
images of the +15- and +30-v composite curves. Note that the slope of 
the composite characteristics is about twice as great as the slope of the 
individual characteristic at the Q point, thus verifying thatTj, (composite) 
= rJ2. 

Observe that the quiescent operating point for the composite tube of 
Fig. 10.20 is at v P = 300 v and i P = 0. The a-c load line for the com- 
posite tube is drawn through this Q point. The slope of this load line 
should be determined from power output, power gain, and plate dissipation 
considerations in the same manner as discussed for the single tube. The 
slope of the composite load line for a triode can frequently be made the 
negative of the slope of the composite characteristic, thus providing 
maximum power gain at maximum power output. This impedance 
matching provides somewhat more than twice the power output for the 
composite tube than could be obtained from the individual tube at 
acceptable distortion levels. The slope of the composite load line represents 
a resistance which is one-fourth of the plate-to-plate resistance of the 
output transformer, because the composite amplifier works into one-half 
of the output transformer. 

As shown in Fig. 10.20, the operating path of each tube in the push-pull 
amplifier is a curved path which passes through the individual Q point. 
The composite tube current is 

'pc = hi ~ ip2 (10.46) 



Large-Signal Amplifiers 



409 



-1.0 V 



-0.8 



-10 



Vet in volts 
-15 -20 



-25 



-30 



8. 

E 



-0.6 



,-0.4 



-0.2 




-30 



-25 



-20 -15 

V c e in volts 



Fig. 10.21. Composite characteristics and load lines for push-pull transistors. 

Therefore, the plate current in the individual tube may be determined at a 
particular value of grid voltage from the relationship 

»« = tpc + ip^ (10.47) 

Then, if a vertical line is drawn through the intersection of a composite 
characteristic with the load line, the intersection of this vertical line with 
the corresponding individual tube characteristics will provide a point on 
the operating line for the individual tube. This construction is illustrated 
by the points a, b, and c on Fig. 10.20. 

Figure 10.21 shows that the construction techniques illustrated for the 
push-pull triode amplifier also apply to the push-pull transistor amplifier. 



410 Electronic Engineering 

Since the plate characteristics of the pentode tube and the transistor 
collector characteristics are very similarly shaped, the composite char- 
acteristic construction of a pentode amplifier follows that of the transistor 
very closely. The main difference between these two amplifiers and the 
triode amplifier is the relative slopes of their individual characteristics. 

PROB. 10.13. Two triode-connected 6L6 tubes are used in a push-pull amplifier 
circuit. The quiescent plate-cathode voltage is 300 v. 

(a) Draw a circuit diagram for the amplifier, using cathode bias. 

(b) Construct the composite plate characteristics, using V G = — 30 v as the Q 
point of the individual tube. 

(c) Draw a suitable load line. What plate-to-plate resistance does this load 
line represent ? 

(d) Draw the operating line for the individual tube. 

(e) Calculate the maximum power output obtainable when the excitation is 
sinusoidal. 

(/) What should be the value of plate-supply voltage for this amplifier ? 
PROB. 10.14. Two 2N2147 transistors are used in a push-pull amplifier circuit 
with V cc = — 20 v and I B = —10 ma at the individual Q point. Sketch the 
composite characteristics and draw an appropriate load line. What value of load 
resistance does this represent ? Determine the maximum available power output. 
Assume that the exciting source is sinusoidal and the source resistance is large in 
comparison with h ie . 

10.6 CLASS B OR AB OPERATION 

Almost all the amplifiers considered in the preceding portion of this 
text have been class A amplifiers. Class A amplifiers were used because it 
was assumed that linearity, or minimum waveform distortion, was a 
requirement of the amplifier. From the definition of the classes of opera- 
tion, it is evident that class A is the only class of operation which can 
provide good linearity in a single-ended (non push-pull) audio ampli- 
fier. However, the even-harmonic cancelling property of a push-pull 
amplifier permits class B or AB operation for this type of amplifier. 

If the amplifying device, such as a tube or transistor, were linear during 
the time current flowed through the device, the push-pull amplifier would 
be linear even though the individual amplifier were operated class B 
(biased at cutoff). In this case, the output current and power would be 
furnished by one of the individual amplifiers during the first half of the 
input cycle and the other individual amplifier would furnish the output 
current and power during the second half of the input cycle. The addition 
of the two individual currents would produce a distortionless output cycle. 

The no-signal output current for a class B amplifier is zero. Therefore, 
when the excitation is zero, there is no drain on the collector or plate 
power supply. This feature of the class B amplifier is very attractive 



411 

Large-Signal Amplifiers 

because of the large reduction of collector supply power in the class B 
amplifier compared with the class A amplifier when equal output power 
capabilities are considered. 

In addition to the greatly increased small-signal efficiency, the class B 
amplifier is considerably more efficient than the class A amplifier at 
maximum signal levels. The maximum theoretical efficiency of the class B 
amplifier, for sinusoidal excitation, may be readily computed by assuming 
the amplifier operation to extend from the voltage axis along the load line 
to the current axis. This technique was used to calculate the maximum 
theoretical efficiency of the class A amplifier, which is 50% with sinusoidal 

excitation. 

Load lines are drawn on the output (collector or plate) characteristics of 
typical transistor and tube amplifiers in Fig. 10.22. When the operation 
is assumed to swing from axis to axis along the load line, the peak value of 
input current is J A . Now, the average current for one half of a sine wave 
with a peak value of I A is (2/tt)/^. Thus, if the transistor is used as an 
example the collector power input P ln is equal to V cc times the average, 
collector current or 

Pin = V^oc (10 .48) 

77 

where I A is the current axis intercept value of i c . This Pin is the average 
power into one transistor for one half-cycle or for two push-pull transistors 
for the full cycle. For the push-pull configuration, the characteristics of 
Fig. 10.22 represent the top half of the composite characteristics. Here, the 
average power output for either the half cycle or full cycle is 

Pout = lA^£ (10.49) 

The maximum theoretical collector circuit efficiency is 

„ *22* x 100 = - x 100 = 78% (10.50) 

/C Pin 4 

This value of efficiency is closely approached by the transistor amplifier 
and reasonably approached by the pentode amplifier. It is interesting to 
note that the triode amplifier is quite inefficient, compared with the other 
two, when the maximum permissible grid voltage is zero volts. However, 
when the grid is driven appreciably into the positive region, the efficiency 
of the triode may be as high as that of the pentode amplifier. 

The collector or plate dissipation capability of an amplifier usually 
limits the amount of power that can be obtained from this amplifier. 
Thus, a given amplifier can deliver much more power when operated in the 



412 



Electronic Engineering 



-2.5 



-2.0 - 



t-1.5 



■g -1.0 



-0.5 



1 1 1 1 

25. 


1 




\s — """ 2 ° 






]V — js- 




- 


-v^"\"^ ^2 




- 


i i i N 


5 P* 


- 



250 



200 - 



1 150 

E 



- « 100 - 



50 



-2 -4 -6 -8 -10 -12 

Collector volts 

(a) 



1 


1 1 


1 


i 


— f 






- 


\ 







- 


1 1 


i i 


r 


-15 v - 

7?- 



100 200 300 400 500 600 
Plate volts 
(b) 



200 






1 


i 


I i 




* 150 

E 

I 
a 100 

ro 
a. 




"O/ 

— »/ 


'v 

*/ . 

/ W 

/ */ 

(^7 




/ ' 


- 


50 










^ 


— 













**f** , 





100 



200 300 400 

Plate volts 

(c) 



500 600 



Fig. 10.22. Load lines are drawn on the output characteristics to determine the max- 
imum theoretical efficiency and to compare the efficiency of transistor and tube am- 
plifiers under various situations, (a) Transistor; (b) pentode; (c) triode. 



Large-Signal Amplifiers 413 

class B mode than when operated in the class A mode. The relationship 
between the maximum possible power output and the efficiency is derived 
as follows. The collector or plate dissipation is 

P d = Pm - Pout (10.51) 

But 

P ln = ^ (10.52) 

Vc 
Then 

P d = Pout U - l) (10.53) 

and 

P out = JsEi- (10.54) 

1 ~Vc 

PROB. 10.15. A certain transistor can safely dissipate 10 w in a given environ- 
ment. Determine the power output which can be obtained from a push-pull 
amplifier using two of these transistors when the transistors are operated (a) 
class A, (b) class B. Assume the excitation to be sinusoidal and the efficiencies 
to be 50% and 75 % respectively. 

As in class A operation, the class B efficiency is a function of waveform 
as well as magnitude of excitation. In fact, as mentioned, the efficiency 
may approach 100% when the excitation is rectangular. 

The realizable increased power output of the class B mode as compared 
with the class A mode is not obtained automatically as the bias is changed. 
This maximum power output can result only from increased collector 
power input. The collector power input can be increased by either 
increasing the collector supply voltage or increasing the output current or a 
combination of the two. In case the collector supply voltage has already 
been set at the maximum permissible value, the output current can be 
increased by reducing the load resistance and increasing the excitation. 
For the vacuum tube, this means that to realize the maximum dissipation 
capabilities of the tube the grid must be driven into the positive region. 
The triode tube looks quite promising when driven into the positive grid 
voltage region because the efficiency as well as the power output, is then 
increased appreciably. 

Actually, either the tube or the transistor push-pull amplifier has 
intolerable distortion when biased strictly at cutoff. This high distortion 
results from the high degree of nonlinearity of the characteristics near zero 
collector or plate current. This high distortion which occurs near plate 
current cutoff is known as cross-over distortion. This distortion can be 
eliminated by biasing the amplifier at projected cutoff rather than at actual 



414 



Electronic Engineering 



cutoff. The meaning of projected cutoff is illustrated by Fig. 10.23. The 
relatively straight portion of the transfer or input characteristic curve is 
extended until it intercepts the zero plate current or base current line as the 
case may be. This intersection is at the projected cutoff value of bias 
voltage as shown in Fig. 10.23a. The composite transfer characteristic 
may be drawn by inverting the second transfer curve and aligning the 
projected cutoff bias points as shown in Fig. 10.23a. 

The composite transistor input characteristic can be drawn for the case 
of projected cutoff bias, as shown in Fig. 10.236. The second set of 



*B2 



-60 -30 




10 5 



lj>2 U B2 




/„ 5 10 



l Bl 



(a) 



(b) 



Fig. 10.23. Illustration of the method of determining projected cutoff for the (a) 
vacuum tube, (b) transistor. 



characteristics is inverted and the projected cutoff bias voltage values of 
the two curves is aligned. The extensions of the relatively straight portions 
of the curves then join and form the composite characteristic. The 
addition of the second curve is really unnecessary because no additional 
information is added. The extended linear portion of the curve becomes 
the composite characteristic for one-half of the input cycle and the second 
half cycle is merely a repetition of the first half cycle if the amplifiers are 
balanced. 

Operation at projected cutoff is sometimes called class B operation in 
the literature. However, it is more correctly known as class AB operation. 
The efficiency of class AB is less than that of class B, of course, but is still 
much higher than that of class A . Push-pull amplifiers are usually operated 
at projected cutoff. The chief disadvantage of this mode of operation in 
comparison with class A is the increased importance of maintaining good 
amplifier balance. 

The composite characteristics and composite load line can be drawn 
for the class AB push-pull amplifier in the manner described for the class A 



415 

Large-Signal Amplifiers 

amplifier The only difference in the two amplifiers is the selection of the 
bias point. However, the construction of the composite curves is much 
easier for the class AB amplifier because the composite curves follow 
the individual curves over most of the region of operation. In fact, the 
composite curves are not required for a class AB amplifier because the 
composite quiescent operating point is fixed independently of the com- 
posite curves, and the composite curves are identical with the individual 
characteristic curves at the extremities of the region of operation. Also, 
as previously mentioned, one set of collector characteristics is sufficient to 



Composite load line 




Composite 
" load line 



/» = 



"ce 



(a) 




Individual 
'operating line 



Fig. 10.24. Graphical analysis of the class AB push-pull amplifier, (a) Transistor; 
(b) triode tube. 

define the operation over one-half of the output cycle, and the second half 
of the output cycle is assumed to be identical (although inverted) to the 
first half. Therefore, only one set of output characteristics needs to be 
used. These simplifications in the graphical analysis are illustrated in 
Fig. 10.24. The maximum positive grid voltage of a vacuum tube operating 
in the class 1 (no grid current) condition is established at zero volts. 
Then the optimum composite load line can be drawn, using the guiding 
principles discussed for the single ended amplifier. However, the maximum 
permissible excursion of either input voltage or output current is not so 
clearly defined for the class AB 2 tube amplifier or for the class AB transistor 
amplifier. The maximum permissible output current can be determined as 
a function of the permissible plate or collector dissipation and the quiescent 
value of output current. This maximum permissible value of output 
current then establishes the upper extremity or current axis intercept of 

the load line. 

A method of determining the maximum permissible output current for 
a class B transistor amplifier is illustrated in Fig. 10.25. Initially, the 



416 



Electronic Engineering 



amplifier is assumed to operate class B so the quiescent current is approxi- 
mately zero. Then with sinusoidal excitation, the average power dissipation 
for one half-cycle is 

Pa = Pm - Pout = - I m * x V cc - ^^ (10.55) 

IT 2 

The value of signal current 7 ma x which causes maximum dissipation may 



I -2.T 




Fig. 10.25. Illustration of the method used to determine the maximum permissible 
collector current for a class B transistor. 

be obtained by differentiating Eq. 10.55 with respect to 7 ma x and equating 
the derivative to zero. 



dl, 



_ \j _ v cc l m& x K L 



(10.56) 



2 V, 



cc 



J z. • w - I 

'max — — — ij 



(10.57) 



Substituting this value of / max into Eq. 10.55 will yield the maximum 
power dissipation. 



4 2 

Pd max = — Ia^CC \^A &1 



(10.58) 



Large-Signal Amplifiers 

Substituting I A R L = V cc into Eq. 10.58 we have 



Pdmax = - : IaYcC 



417 



(10.59) 



For purposes of comparison, sketches of power dissipation and power 
output as functions of /max are given in Fig. 10.26 for both class B and 
class A. 

Equation 10.59 will determine the value of I A for a class B amplifier 
when V cc (or V PP ) and the maximum permissible dissipation are known. 



7T 2 P d 



5P d 



T _ fimax _ Tt fjmax ^ ->* dmax /-jq ^q-v 

A ~ (2/7r 2 )F cc ~ 2V CC ~ V cc 

This equation was derived for one transistor operating over one half- 
cycle. However, when a complete cycle is considered, the average power 



\i A v c 



:IaVc 



— ^ 


y^^\ 



$Ia 



1a 




'max ^ 

Class B 



L max *■ 

Class A 



Fig. 10.26. Power output and power dissipation as functions of / max (sinusoidal signal). 

input and the average power output remain the same as in the preceding 
equations, but the power dissipation is the total dissipation capability of 
the two transistors (or tubes). A more general equation which would be 
valid for class AB operation, and thus include I Q , would be more useful 
than Eq. 10.60. Such an equation is difficult to derive. Equation 10.60 
can be altered, however, to include I Q and to provide approximate values 
of I A for class AB (or even class A) operations. The altered equation is 



5P a 



-3/, 



(10.61) 



cc 



Although the foregoing relationships were developed for sinusoidal 
signals, they are also safe design values for random signals. 



418 Electronic Engineering 

PROB. 10.16. A particular transistor can safely dissipate 5 w with a given heat 
sink. The projected cutoff bias of this transistor provides a quiescent collector 
current of 20 ma. Calculate the maximum permissible collector current I A , the 
collector-to-collector load resistance, and the maximum power output when two 
of these transistors are operated in class AB push-pull. The excitation is 
sinusoidal and V cc is 15 v. Answer: l A = 3.27 a, R L = 18.4 ft, P = 24 w. 

10.7 THERMAL CONDUCTION AND THERMAL RUNAWAY 

As noted in Chapter 4, small-signal transistors have collector-junction 
dissipation ratings which are based on an ambient air temperature, usually 
25°C. These transistors, like small (receiving) tubes rely on air convection 
to transfer the heat, which results from the dissipated power, to the 
surrounding atmosphere. The small-signal transistors must be derated as 
the ambient temperature rises above the reference temperature. Convec- 
tion-cooled tubes should also be derated as the ambient temperature 
increases, but the normal changes of ambient temperature are usually 
negligible in comparison with the operating temperature of the tube. 
Therefore, little attention is usually given to derating vacuum tubes, 
although the need for derating should not be overlooked. 

Forced air (or even water) cooling is frequently recommended by the 
manufacturer for high-power tubes such as transmitting tubes. In contrast, 
the heat from a power transistor is usually transferred by conduction to a 
heat sink. Therefore, the power dissipation rating of a power transistor is 
specified for a given case temperature, usually 25°C. The actual power 
which can be dissipated by the transistor depends on the thermal conduc- 
tivity between the transistor case and the surrounding air or environment, 
as well as on the ambient temperature of the surroundings. The power 
transistor (500 mw and above) is usually fastened securely to a large metal 
plate or chassis, which serves as the heat sink. The collector of the power 
transistor is usually electrically and mechanically connected to the tran- 
sistor case. Maximum thermal conductivity to the heat sink is obtained 
when the transistor case is electrically and mechanically fastened to the 
heat sink. However, an insulator is frequently required to electrically 
isolate the collector from the heat sink. 

Every transistor has a maximum permissible junction temperature Tj, 
which ranges from 85 to 110°C for germanium transistors and is about 
1 75°C for silicon transistors. Thermal resistance ® T has been defined as the 
ratio of temperature rise in degrees centigrade to the power conducted in 
watts. 3 Therefore, the junction temperature T t can be related to the power 
being dissipated P a , the thermal resistance Q T , and the ambient temperature 

3 The symbol may not seem to be an appropriate symbol for thermal resistance. 
However, this symbol is commonly used by the transistor industry. 



Large-Signal Amplifiers 



419 



(10.62) 



T a by the equation 

T, = @ T P d + T a 

The thermal resistance r actually consists of three parts; the thermal 
resistance between the collector junction and the case 0, c , between the case 
and the heat sink CS , and between the heat sink and the ambient environ- 
ment ©„,. The equivalent circuit of Fig. 10.27 represents the thermal 
relationships of the transistor and its heat sink. The capacitors shown 
represent the thermal capacitance of the three parts of the circuit. The 
thermal capacitance Q of the collector junction is very small. Therefore, 



Collector junction temperature 



< Case temperature T c 




< Heat sink temperature T s 



Ambient temperature T a 



Absolute zero temperature 

Fig. 10.27. An equivalent circuit which relates the various temperatures of a heat- 
conducting system to the power being dissipated and the thermal resistance. 

the thermal time constant <d ie C, is of the order of milliseconds. The 
significance of this time constant is that the load line can cross the maxi- 
mum dissipation line of a transistor without damaging the transistor, 
providing that the excessive dissipation does not continue for more than a few 
milliseconds. The thermal capacitance C c of the case is much greater than 
that of the junction. Therefore, the transistor which is designed to operate 
with a heat sink can operate for at least several seconds without the heat 
sink before the transistor is damaged. Similarly, the thermal capacity of a 
heat sink is usually much larger than that of the transistor case. Therefore, 
the transistor may operate several minutes with an inadequate heat sink. 

The thermal resistance JC from the collector junction to the case is 
usually given by the manufacturer for each type of power transistor. Also, 
thermal resistance data are available for the various transistor mounting 
systems. 4 The thermal resistance SO of a J-in. thick bright aluminum 

1 See Motorola Power Transistor Handbook, First edition, p. 23, Motorola Semi- 
conductor Products Division, Inc., Phoenix, Arizona. 



420 Electronic Engineering 

plate as a function of area (both sides) is given in Fig. 10.28. This chart is 
for vertical orientation. When the heat sink is horizontal, the thermal 
resistance increases by the order of 10 % because of the reduced convection. 
On the other hand, black painting or anodizing of the heat sink could 
lower its thermal resistance. 



1000 



~ 100 



10 













i i i i 
























square sheets of )g-inch thick 
bright aluminum. Heat sinks w 


































ere 
















held vertically in still air. (Heat sink 














area is twice the area of one side) 








































Exp 
' 9 


erimental average 
sa = 32.6A-- 472 




























































































































































































































' 


\ 















10 
B SA , thermal resistance ("C/watt) 



100 



Fig. 10.28. Heat sink area vs. thermal resistance. (Courtesy Motorola Semiconductor 
Products Division.) 

PROB. 10.17. A particular power transistor which has & jc = 0.5°C/w is 
mounted on a vertical |-in. aluminum 200-in. 2 heat sink. The maximum junction 
temperature is 90°C and the mounting system has ©„, = 0.8°C/w. How much 
power can be dissipated when the ambient temperature is 25°C? What is the 
dissipation rating of the transistor? 

The transistor is not completely protected when the heat sink is adequate 
for the required dissipation; the transistor can be destroyed by thermal 
runaway. This thermal runaway results from the increasing collector 
current, caused by the increasing I co which occurs as the junction tempera- 
ture rises. Thus, thermal stability is dependent on the current stability 
factor S z as well as on the heat sink. 

Thermal runaway occurs when the rate of increase of collector junction 
temperature exceeds the ability of the heat sink to remove the heat. The 
rate of increase of junction temperature is obtained by differentiating Eq. 
10.62, 

dT, = r dP d (10.63) 

where ® T = @ ic + @ cs + @ OT . At quiescent operating conditions, the 



Large-Signal Amplifiers 421 

power dissipation P d is equal to the collector power input V C I C . Then 

3T, = @ T d(V c I c ) = @ T V C dl c (10.64) 



dT t _ 
Then, 



^ = V C S T (10.65) 

dl c 



But dl c \dl co = S It the current stability factor noted (Chapter 5). 
Thus, 

-^- = SjV c @ T (10.67) 

dlco 
In a germanium transistor, J co theoretically doubles for each 10°C 
temperature increase, as previously mentioned. Then J co increases about 
7 % for each °C increase of junction temperature. This relationship can be 
expressed mathematically as follows: 



dlco 



= 0.07/ co (10.68) 



9T t 

where I co is determined at the operating junction temperature T t . 
Substituting Eq. 10.68 into Eq. 10.67, we have 

007Ic ° = 7ihr (1069) 

O.O7/ co S z K c r = 1 (10.70) 

Equation 10.70 expresses the parameter relationship at the transition 
between the stable condition and thermal runaway. The left side of this 
equation must be less than unity to assure stability. After the collector 
supply voltage V c is set and the heat sink is designed, the maximum value 
of stability factor which will provide thermal stability may be calculated. 

S 7 < 143 (10.71) 

The constant in the numerator of Eq. 10.71 is theoretically appropriate 
for germanium transistors. However, the rate of change of I co with 
temperature varies quite widely among different types of transistors. The 
actual rate of increase is less than the theory predicts. This is especially 
true of silicon transistors because the surface leakage, the carriers gener- 
ated within the depletion region and so on, which account for the discrep- 
ancy, cause a larger portion of I co in the silicon transistor. Therefore the 
constant in the numerator of Eq. 10.71 should be determined for each 



422 Electronic Engineering 

particular transistor. This constant can usually be determined from the 
data furnished by the manufacturer. For example, if I co is given for two 
different temperatures, Eq. 10.72 can be used to find the constant which 
represents the rate of increase of I co with temperature. 

W = Wl + r)^' (10.72) 

where r is the rate of increase, and \\r is the numerator of Eq. 10.71. 

The I co in Eq. 10.71 must be determined at the maximum junction tem- 
perature, which is equal to the maximum ambient temperature plus the 
maximum d-c collector dissipation times the total thermal resistance. The 
maximum collector dissipation is a function of the junction temperature 
because Sj^q is a significant part of the collector current. The maximum 
junction temperature is T s max = T a max + CP in - P m t) d T- Then 

T, max = T a max + (J c + Sjlao max ) V C @ T - P out d T (10.73) 

where I c is the average value of collector current determined at maximum 
dissipation conditions but neglecting I co . Then 

J c V c = T max ~ T amax _ Si y cIcomaK + p Qut (10J4) 

fc) T 

where I C V C is the permissible collector power input with I co neglected. 
When the permissible collector power input is determined from Eq. 
10.74 and the maximum stability factor is determined from Eq. 10.71 
(using T co max ), the junction temperature will not exceed the design value. 
The ohmic resistance in the collector circuit provides a safety factor because 
V c normally decreases as the total collector current increases. The R-C 
coupled amplifier does not present a thermal runaway problem because of 
the comparatively large value of ohmic resistance in the collector circuit. 

PROB. 10.18. Two transistors of the the type used in Prob. 10.17 have I co = 
1.5 ma at 25°C and I co = 30 ma at 90°C measured at V c = -30 v. What is the 
maximum permissible stability factor for a transformer coupled amplifier with 
V c = — 30 v, using these transistors? Both transistors are mounted on a 
^-in thick aluminum heat sink of 200 in 2 area on one side. Answer: Sj = 5.24. 
PROB. 10.19. The transistor amplifier of Prob. 10.18 is operated class AB. If 
I c max and I c ave are determined with the aid of the typical collector character- 
istics, as discussed in the preceding problem, what is the maximum permissible 
value of I c ave ? 

10.8 PHASE INVERTER CIRCUITS 

In some instances, a transformer may be undesirable in the input of a 
push-pull amplifier. For this reason, phase inverter circuits which provide 
balanced voltages of opposite polarity have been devised. One rather 



Large-Signal Amplifiers 423 

simple type of phase inverter circuit is shown in Fig. 10.29. This phase 
inverter makes use of the inversion characteristics of a conventional 
amplifier. The tube V x inverts the signal of grid G x and applies this inverted 
signal to the grid G 2 . The tap on resistor R gl is adjusted until the magni- 
tudes of the signals on G 1 and G 2 are equal. In fact, this circuit can even 








Push-pull 
stage 



Fig. 10.29. A simple type of phase inverter circuit. 

be adjusted to compensate for differences in the amplification of the 
push-pull tubes. 

PROB. 10.20. A vacuum tube amplifier is to be used in the circuit of Fig. 10.29. 
If R gl = R g2 = 500 Kii and R L2 is 50 Kfi, find where the tap on R gl must be 
located. Assume g m of the triode is 2000 /tmhos and r P of the triode is 10,000 n. 
Answer: 30.5 KCl from ground. 

A transistor counterpart exists for the phase inverter circuit of Fig. 10.29. 
This transistor phase inverter is shown in Fig. 10.30. This circuit contains 
stabilized bias in each stage. Accordingly, a large number of circuit 
elements are required. However, by the proper choice of bias currents, 
several of the circuit elements may be eliminated. For example, if the bias 
current to transistor 7\ is less than the bias current of transistor T 2 , 
resistors R 3 and R 2 may be removed and capacitor Q replaced by a short 
circuit. Bias for transistor 7\ is supplied by R s and R 7 . Obviously, the 
adjustment of R x controls the output current of transistor T x . 

The greater complexity of the transistor circuit (Fig. 10.30) as compared 
to the vacuum tube circuit (Fig. 10.29) results from the bias requirements. 
In vacuum tube circuits, the bias circuit can be located in the cathode of 
each tube. Hence, isolation of the grid circuits are not required. In 
contrast, the bias circuits of transistor amplifiers are located in the base 
circuits. Consequently, isolation of the base circuits must be maintained. 



424 



Electronic Engineering 




Input 



Fig. 10.30. A transistor phase inverter circuit. 

PROB. 10.21. A transistor circuit is connected as shown in Fig. 10.30. All 
three transistors are the same type (transistor 7\ is usually a lower current rating 
device than transistors T 2 and T^). h ie = 250 fl; h re r^0; h oe — 10 _5 mhos; 
h fe or P = 50. The desired quiescent conditions are I B = 1 ma and I c = 50 ma. 
Let the currents through R 2 , R s , and R 5 be 3 ma each. Now, if R 9 = 10 O and 
R t = 20 ft, find the value of the other circuit elements including R v Assume R 10 
is 300 ft and V cc is 20 v. The lowest frequency to be amplified is 30 Hz. 

A second type of phase inverter circuit which has proven very popular 
is shown in Fig. 10.31. In this circuit, R Ll and the vacuum tube act as a 
cathode follower. Accordingly, the signal in grid G 2 is in phase with the 
signal v { . In contrast, the signal on grid G 1 is developed across the con- 
ventional plate load resistance R L2 and is inverted in relation to the signal 
v { . Hence, if R Ll = R L2 , the two signals will be equal in magnitude (the 
same current flows through both R Ll and R L2 ) and inverted in phase. 

PROB. 10.22. Show how the phase inverter of Fig. 10.31 can be directly coupled 
to the plate of the amplifier which precedes it. Use triodes for both the phase 
inverter and the preceding amplifier. A single duo-triode would serve both 
functions. 

Since the cathode circuit acts as a cathode follower, the total voltage 
gain in the cathode circuit is less than one. Similarly, since the voltage 
gain in the plate circuit is equal to the voltage gain in the cathode circuit, 



Large-Signal Amplifiers 



425 




Push-pull 
-i [ stage 



Fig. 10.31. A split-load phase inverter circuit. 

the total voltage gain in the plate circuit is less than one. However, since 
the voltage in the cathode is inverted in relation to the voltage in the 
plate, the grid-to-grid voltage on the push-pull stages may have a gain 
greater than one (but never greater than two). 

The transistor version of the phase inverter of Fig. 10.31 is shown in 
Fig. 10.32. The operation of this circuit is the same as for a vacuum tube. 
However, in this version, the currents flowing in the base circuits of T 2 and 
T 3 must be equal. As for the vacuum tube, the voltage across R L1 will be 
equal to the voltage across R L2 if transistors T 2 and T 3 are not present. 5 




Input 



Fig. 10.32. The transistor version of Fig. 10.3J 



5 R L1 passes a slightly greater current (I c + I B ) than the current through /?«, (I c ). 
This current difference is usually negligible but can be compensated for by making R L1 
less than R Ll . 



426 Electronic Engineering 

When the push-pull transistors are connected to the phase inverter, 
the voltages across the emitter and collector impedances will be equal and 
of good waveform provided that these impedances remain equal. The total 
emitter current is approximately equal to the input voltage divided by the 
total emitter circuit impedance. The collector current is essentially equal 
to the emitter current. If the emitter circuit impedance varies over the 
signal cycle, the emitter current and hence the collector current will not be 
proportional to the input voltage. The emitter voltage will be approxi- 
mately equal to the input voltage, as mentioned, but the collector voltage 
will not have the same waveform as the input voltage unless the collector 
circuit impedance varies in the same manner and in phase with the emitter 
circuit impedance. This is too much to expect of the collector circuit 
impedance, since the push-pull transistors must have opposite polarities. 
Therefore, the emitter and collector circuit impedances must be essentially 
linear, or nonvarying. 

The load impedances of the split load phase inverter can be linearized 
by making the shunt fixed resistances, such as R Ll and R L2 , small in com- 
parison with the input resistance to the transistor. Linearization can also 
be accomplished by placing fixed resistance in series with the transistor 
input resistance. This method is used in the circuit of Fig. 10.32. A much 
smaller resistance in the emitter circuits of the push-pull transistors would 
accomplish the same purpose, providing that they are not bypassed. 

The two diodes shown in the base circuits of transistors T 2 and T 3 
(Fig. 10.32) are needed when the push-pull amplifier is operated class AB 
or class B, because the base current of the power transistors tend to charge 
the coupling capacitors C cl and C C2 during the portion of the input cycle 
when the particular transistor is conducting. However, the coupling 
capacitors cannot discharge through the transistor base during that portion 
of the cycle when the transistor is cut off. Therefore, since the coupling 
capacitor discharge time through the bias resistors is large in comparison 
with a period, the coupling capacitors acquire a net charge which tends to 
reverse bias the transistors (class C). The diodes which are placed in the 
base circuits will prevent this shift of operating pointby providing a dis- 
charge time constant which is approximately equal to the charging time 
constant. 

The voltage gain of a split load phase inverter is always less than two, 
but the current gain can be large. Accordingly, considerable power gain 
is possible. Other phase inverter circuits are, of course, possible. The 
reader is referred to the various handbooks for these other circuits. 

PROB. 10.23. A circuit is connected as shown in Fig. 10.31. (a) If Jt L1 = R L2 = 
5000 £2, what is the grid-to-grid voltage if v t is a sine wave with 1 v peak. Assume 
g m is 5000 /<mhos and r p is 10,000 £2. (6) How would you modify the circuit if 



Large-Signal Amplifiers 

the desired quiescent point for the phase inverter tube is V G = -4 v when 
J p = 8 ma? Answer: (a) 1.85 v peak. 

PROB. 10.24. A circuit is connected as shown in Fig. 10.32. (a) Find i b2 if the 
signal /„ is 1 ma peak. Assume R L1 = R L2 = 1000 Q. The input impedances of 
J, and r, are each 200 Q. The characteristics of transistor 7\ are h ie = 250 Q, 
h re = 0, h oe = 10 micromhos, h u or P = 50. * 2 and * 3 are much larger than 
200 O. (b) What value should R L1 have if R L2 is 1000 Ci and perfect balance is 
desired? Assume tf 5 = R e = 500 Q and R t = R t = 1000 n. 

10.9 GRID DRIVE REQUIREMENTS FOR CLASS B 2 (or AB 2 ) 
VACUUM TUBE OPERATION 

A rather drastic change occurs in the control grid circuit of a tube when 
the control grid is driven positive. The positive control grid intercepts 
electrons from the cathode and current flows in the control grid circuit. 




2 3 

Grid volts 

Fig. 10.33. Control-grid current as a function of positive control grid voltage. 

The curves of control grid current vs the control grid potential are given in 
Fig. 10.33. These curves indicate that the control grid current follows the 
same pattern as does plate current in a diode. In contrast to a diode, the 
triode plate voltage does have an effect on grid current, but this effect is 
quite small. 6 Hence, the positive control grid can be roughly represented 
as a resistor from grid to cathode. The size of this resistor is usually in 
the order of 1000 Q. or less. 

As a result of the action of the positive control grid, the input character- 
istics of a push-pull class B 2 (or AB 2 ) vacuum tube amplifier has the form 
indicated in Fig. 10.34. The voltage A0 (and B0) represents the bias 

6 Plate voltage can have a big effect on grid current if the plate voltage approaches 
the magnitude of the control grid voltage. 



428 



Electronic Engineering 



A + 

Voltage input 



Fig. 10.34. Input characteristics of a class B 2 (or AB 2 ) vacuum tube push-pull amplifier. 

potential of the push-pull grids. As soon as this bias potential is exceeded, 
the control grid begins to draw current. This control grid current causes 
a load to be applied to the driver stage which supplies the signal voltage 
for the push-pull tubes. As a result of this load, which is only present for the 
peaks of the input signal, the signal from a high internal impedance driver 
stage is distorted as shown in Fig. 10.35. The symmetrical appearance of 
the distorted waveform indicates that odd harmonics have been introduced 
into the waveform (Fig. 10.5). Usually, the magnitude of the third 
harmonic component is much larger than the other odd harmonics. Thus, 
Eq. 10.22 can be used to find the peak value of the third harmonic 
component, V 3 . 



V a = 



2V w/6 - V m 



(10.75) 



where K max , V Q , and K 6 have the values shown in Fig. 10.35. Similarly, 
Eq. 10.21 may be used to obtain the peak magnitude of the fundamental 
component of grid voltage V x . 



V x = 



2V m 



+ 2V r/t - 4V 



(10.76) 




Fig. 10.35. Distortion in the input signal of a class B 2 (or AB t ) push-pull vacuum 
tube amplifier. 



Large-Signal Amplifiers 429 

The foregoing analysis reveals the special requirements a driver stage 
must have to properly operate a class B 2 (or AB 2 ) stage. 

1 . The driver stage must be capable of supplying the power required by the 
positive control grids of the push-pull stage. 

2. The output impedance of the driver stage must be much less than the 
minimum grid circuit impedance for low distortion. 

The first condition determines the required size of driver tube. The 
second condition determines the type of circuit which produces least 
distortion. Because of the low value of positive grid-to-cathode impedance, 
transformer coupling is almost always used. (The cathode follower could 
also be used as a push-pull driver.) This transformer coupling allows the 
high output impedance of the driver stage to be lowered to meet the low 
grid-cathode impedance requirement. In addition, the phase inversion for 
the push-pull grids is achieved by the transformer. 

PROB. 10.25. The output impedance of a driver tube is 10,000 CI. This driver 
tube is coupled through a transformer to a set of push-pull tubes which have a 
bias voltage of -25 v. The transformer has 500 turns in the driver plate winding 
and 1000 turns (center tapped) in the grid-to-grid circuit. The positive grid-to- 
cathode impedance of the push-pull tubes is approximately 1000 CI. The magni- 
tude of the signal from the driver stage is 35 v peak when the push-pull tubes are 
removed from the circuit, (a) What is the peak value of driver voltage with the 
push-pull tubes in the circuit? (6) What percentage of third harmonic signal is 
present under these conditions ? (c) What is the maximum output power required 
in the driver stage? Answer: (a) 25.9 v, (6) 10.5%, and (c) 23.3 mw peak inst. 
power. 

PROB. 10.26. Repeat Prob. 10.25 if the signal out of the driver tube is tripled 
and the number of turns on the driver side of the transformer is tripled (1500 
turns). The number of turns on the grid side of the transformer are the same as 
in Prob. 10.25. 

10.10 COMPLEMENTARY SYMMETRY AMPLIFIERS 

Because of the availability of both p-n-p and n-p-n configurations, the 
transistor allows more versatility of design than does the vacuum tube. 
One interesting circuit arrangement that does not have a vacuum tube 
counterpart is the complementary symmetry amplifier. This amplifier 
employs one p-n-p and one n-p-n transistor of similar characteristics in a 
push-pull arrangement as shown in Fig. 10.36. The two transistors are 
actually driven in parallel, thus no phase inverter is needed. When the 
transistors are biased at projected cutoff, the resistor R t is very small in 
comparison with the resistors R 2 and R 3 . Therefore, /?! does not materially 
increase the input impedance of the amplifier, even though it is not by- 
passed. The resistor Ri may be a potentiometer and thus be used to 
improve the balance of the amplifier. 



430 



Electronic Engineering 




Input 



Fig. 10.36. A complementary symmetry push-pull amplifier. 

Observe from Fig. 10.36 that the positions of the load resistance and the 
power supply have been reversed in the complementary symmetry amplifier 
as compared with the conventional push-pull amplifier. Thus, the signal 
currents add in the load, but the d-c currents do not pass through the load 
when the load is placed in the common branch. Therefore, no output 
transformer is required in this push-pull amplifier. However, a transformer 
may be needed for impedance matching when the available load is not a 
suitable impedance for the transistors. Fortunately, the popular loud- 
speaker impedances are suitable as loads for power transistors. 

An A-parameter equivalent circuit of the complementary symmetry 
amplifier is given in Fig. 10.37. Inspection of this circuit shows that the 




Fig. 10.37. An A-parameter circuit for the complementary symmetry amplifier of 
Fig. 10.36. 



Large-Signal Amplifiers 



431 



composite A-parameter circuit of Fig. 10.38a is also valid. In both of these 
circuits, class A operation is assumed. In fact, the equivalent circuit of 
Fig. 10.37 is valid only for class A operation because the individual tran- 
sistors must operate linearly for the h parameters to be meaningful. 
However, the composite equivalent circuit may be applicable to class AB 
operation as well as class A, providing that the composite characteristics 
are linear. The composite h parameters for class AB operation are 
equivalent to the class A, h parameters of the individual transistor, as 
previously discussed, because essentially only one transistor is conducting 
during each half cycle. The class AB /j-parameter circuit is shown in Fig. 
10.386. Observe that the actual load resistance R L is the composite load 





(a) (b) 

Fig. 10.38. The A-parameter equivalent circuit of the composite transistor (a) class A, 
and (b) class AB. 

resistance. Therefore, the composite load line has a slope equal to the 
reciprocal of R L not RJ4 as in the conventional push-pull amplifier. 
Composite characteristics may be obtained and graphical analysis per- 
formed for the complementary symmetry amplifier in the same manner as 
for the conventional push-pull amplifier. 

The possible elimination of the output transformer with its cost, weight, 
and response characteristic disadvantages are very enticing in favor of the 
complementary symmetry amplifier. However, this circuit also has 
disadvantages. First, two power supplies are needed, as shown in Fig. 
10.36. Also, in the common emitter circuit of Fig. 10.36 the power supplies 
cannot be at ground signal potential because the output signal voltage 
appears between the power supplies and ground. The shunt capacitance 
which exists between the power supplies and ground makes this circuit 
undesirable. Also, the power supplies are not usable for other amplifiers 
in the system unless a common ground exists. 

A circuit which places the power supplies at signal ground potential is 
shown in Fig. 10.39. This circuit differs only from the common emitter 
circuit of Fig. 10.36 in that the ground or common point has been moved 



432 



Electronic Engineering 



to the power supply end of the load resistor. However, this change in 
ground point changes the amplifier to the common collector configuration. 
In addition to the desirable ground point, this configuration provides very 
good linearity because the load voltage is almost identical to the input 
voltage. However, this common collector configuration has high-input 
impedance and for this reason requires comparatively high-voltage 
excitation, thus reducing the power gain and increasing the excitation 
problems. 




Fig. 10.39. A common collector configuration of the complementary symmetry am- 
plifier. 

A final disadvantage of the complementary symmetry configuration is 
the limited availability and high cost of complementary symmetry pairs of 
transistors. A circuit which provides the advantages of the complementary 
circuit but employs two power transistors of the same type is shown in 
Fig. 10.40. This circuit is known as a quasi-complementary circuit. The 
main difference between this circuit and the complementary symmetry 
circuit is the input circuit. Although the power transistor T t is in the 
common emitter configuration, the input impedance of the n-p-n driver 
T 2 matches that of the p-n-p driver 7\ because the effective resistance in 
the emitter circuit of the n-p-n driver is Rz^lI'es aI *d is therefore 
(hf e t + 1)Rl- Although a complementary symmetry pair of transistors 
is used to drive the power transistors, these transistors do not need to be 
closely matched. Therefore, there is little difficulty in obtaining this 
complementary pair. Both driver transistors operate in the common 
collector mode. They have very high (and about equal) input impedances. 

Note that diodes are not needed in the base circuit of the comple- 
mentary symmetry or quasi-complementary amplifiers included here 
because no coupling capacitors are required. When the thermal stability 



Large-Signal Amplifiers 



433 




Fig. 10.40. A quasi-complementary circuit which uses power transistors of the same 
type- 
requirements are stringent, it may become necessary to place thermistors 
or reversed biased diodes in the base circuits, as will be discussed, in order 
to provide adequate thermal stability. 

A quasi-complementary circuit which uses a single power supply is 
shown in Fig. 10.41. The capacitor C in series with the load resistor R L 
permits the load resistor to return to one end of the power supply rather 
than to a center point. Also the transistor 7\ has replaced the lower 




Fig. 10.41. A quasi-complementary circuit which uses a single power supply. 



434 Electronic Engineering 

biasing resistor of Fig. 10.40. This transistor provides voltage gain so 
that the input voltage v T may be a small signal of the order of one volt. 
The thermal stability of the circuit of Fig. 10.41 has been improved by the 
addition of the emitter resistors R E , the base circuit resistors R B , and the 
replacement of the inter-base biasing resistors with diodes D x , D 2 and D 3 . 
These diodes must be selected so that the total quiescent voltage drop 
across them is equal to the sum of the V BE drops across transistors T 2 , 
T 3 , and T t plus the I E R E drop in the emitter circuit of T x at quiescent 
conditions. However the g-point emitter current is normally very near 
zero and the resistor R E must be small in comparison with the load 
resistance R L so that the load power will not be decreased appreciably. 
Therefore the I E R E drop may usually be neglected at the low temperature 
Q point. 

As the input signal v t goes positive (Fig. 10.41), the current through 
transistor T x decreases, and the forward bias of transistor T 2 increases. 
When transistor 7\ reaches cutoff, transistors T 2 and T t should reach 
saturation so maximum voltage is available across the load. Since the 
saturation value of base current for T 2 must flow through R c , the value of 
R c should be chosen so that the voltage drop across R c does not exceed 
one or two volts when this maximum base current flows through it. As 
the input voltage vj goes negative and transistor T x is driven toward 
saturation, the base potential of transistor T 3 approaches ground potential 
on the positive side of V cc . The maximum positive potential of T s (and 
thus the maximum voltage across the load) will be limited by the saturation 
voltage of transistor 7\ plus the i E R E1 voltage drop in its emitter circuit. 
Therefore this maximum voltage drop in the emitter circuit should not be 
more than one or two volts. Otherwise the available output power will 
be seriously reduced. The emitter resistor R E1 may be bypassed with a 
resulting increased voltage gain and a decreased linearity. 

The emitter circuit resistor cannot be bypassed in the class AB amplifier 
because the average emitter current is a function of the excitation. There- 
fore, the bias would be a function of the excitation. Thus an emitter 
resistance which would provide projected cutoff bias at low-signal levels 
would provide class C bias at large-signal levels. When the stability 
requirements are stringent, the required stability factor may be improved 
by replacing the base-to-base resistors (/? x in the circuit of Fig. 10.41) 
with a thermistor or diode. The resistance of this thermistor decreases 
with increasing temperature and therefore causes a more rapid reduction 
of base current with increasing temperature than could be had with a 
resistor in this location. Therefore, the thermistor provides better stability 
than the resistor. 

The circuits included in this chapter are intended to be representative, 



Large-Signal Amplifiers 



435 



not all inclusive. Many additional circuits may be found in the literature. 7 

PROB. 10.27. Two 2N1905 transistors are to be used in a quasi-complementary 
circuit with an 8-fi loudspeaker as a load. Select a suitable value for V cc . Draw 
the load line and calculate the maximum sinusoidal power output available. V CE 
max for the 2N1905 is -40 v. Assume the maximum dissipation with the 
available heat system is 10 w per transistor. Use projected cutoff bias. Answer: 
With V cc = -20v,P o = 22.3 w. 

PROB. 10.28. Two 2N2147 power transistors are used in the circuit of Fig. 
10.41. The n-p-n driver is a 2N3705 and the other two transistors are type 
2N3703. The total power supply voltage is 30 v. The 2N2147's are mounted on 
one 9 in. x 12 in. aluminum chassis with transistor sockets which have 
0.6°C/w thermal resistance. Using projected cutoff, determine: 

1 . Minumum permissible load resistance if the maximum ambient temperature 
is 40°C. 

2. Maximum permissible stability factor Sj when an 8 ii load is used (40°C 
maximum ambient). 

3. Determine suitable values of all components with R L = 8 CI. 

4. Determine maximum sinusoidal power output and required driving voltage 
Vi for your amplifier. 

PROB. 10.29. The typical operation for two 6L6 tubes operating in push-pull 
class AB, is listed in the tube manual as: 

Plate voltage = 360 v 

Screen voltage = 270 v 

Grid voltage = -22.5 v 

Peak grid-to-grid voltage = 45 v 

Zero signal plate current = 88 ma (44 ma/tube) 

Maximum signal plate current = 132 ma 

Zero signal screen current = 5 ma (2.5 ma/tube) 

Maximum signal current = 15 ma 

Effective plate-to-plate load resistance = 6600 

Maximum signal power output = 26.5 watts 



r p = 10K 




e= 5 sin tot 



1000 
turns 




turns 2 o 

+ o 1 " ^ 100' 



100 turns 



2 — - 




Fig. 10.42. The configuration for Prob. 10.30. 



7 One excellent circuits handbook is Seymour Schwartz, Ed., Selected Semiconductor 
Circuits Handbook, Wiley, New York, 1960. 



*3<» Electronic Engineering 

Draw the circuit and determine the magnitude of all resistors and capacitors. 
Assume the lowest frequency to be amplified is 50 Hz. Prove the maximum 
signal power output would be 26.5 watts for the conditions listed. 
PROB. 10.30. A circuit is connected as shown in Fig. 10.42. Plot the voltage 
waveform on the grids of the push-pull circuit. If only third harmonic distortion 
is assumed to be present, find the magnitude of third harmonic component. 
Assume the resistance from cathode to positive control grid is 1000 £1. 
PROB. 10.31. A 2N174 transistor has T imax = 100°C and & ic (junction to 
case) = 0.5°C/)v. If the thermal resistance from case to ambient & ca = 0.5°Clw 
also, determine the current stability factor required to prevent thermal runaway 

(a) the transistor is transformer coupled and V cc = 40 v. 

(b) the transistor must dissipate 30 w. 

(c) the maximum ambient temperature is 35°C. 

id) Ico = 3 ma at 25 "C and doubles for each 10°C temperature increment. 
What is the maximum power dissipation rating of this transistor (case tem- 
perature 25 °Q? 



11 



Amplifiers with Negative 
Feedback 



The performance characteristics of amplifiers may be altered by the use 
of feedback, that is, by adding part or all of the output signal to the input 
signal. If there are an even number of polarity reversals (or no polarity 
reversals) in the closed loop 1 so the feedback signal re-enforces the input 
signal in the mid-frequency range, the feedback is said to be positive. This 
type of feedback is usually employed in oscillator circuits, which are treated 
in Chapter 13. On the other hand, if there are an odd number of polarity 
reversals in the closed loop so the feedback signal tends to cancel the input 
signal in the mid-frequency range, the feedback is said to be negative. This 
negative feedback is the subject of this chapter. Feedback affects the 
driving point impedances as well as the transfer function of the amplifier. 
These effects are considered in the following paragraphs. 

1 The closed loop is the signal path from the input through the amplifier and the feed- 
back system back to the input. 

437 



438 



Electronic Engineering 



11.1 THE EFFECTS OF FEEDBACK ON GAIN, DISTORTION. 
AND NOISE 

The block diagram of an amplifier with negative feedback is given in 
Fig. 11.1. The signal is indicated at the various points by the symbol x. It 
may be seen from the diagram and from the definition of a transfer function 



*i 



K> 



*2 



G23 



*4 



G34 



*3 



Fig. 11.1. The block diagram of an amplifier with negative feedback, 
that the following relations exist. 

x 3 = x 2 G 23 (U.l) 

*4 = x 3 G M (11.2) 

x 2 = x 1 -x i (11.3) 

To obtain the transfer function of the feedback amplifier, x 3 is needed in 
terms of x x . Then, combining the foregoing equations, we have 

x s = (*i - ^)G 23 = (*i - x 3 G 3i )G 23 (1 1.4) 

rearranging terms, 

x 3 (l + G 23 G 3i ) = x x G 23 (11.5) 

and 

X* C/23 



t/ 13 = — = 

x i 1 + G 23 G 3 4 



(11.6) 



or when mid-frequency amplification is considered, the mid-frequency 
reference gain K 13 of the feedback amplifier may be written from Eq. 1 1.6 
as 

where K 23 is the reference gain of the amplifier without feedback and (3 is 
the mid-frequency value of G 3i . This /S is frequently known as the feedback 
factor. 2 In electronic amplifiers, /S is almost always less than unity. In 

2 The symbol fS has been used in previous chapters to represent the current amplifica- 
tion factor of the common emitter transistor. However, /? is also commonly used in the 
literature to represent the feedback factor, determined at mid-frequency. Therefore, /? 
will be used as the feedback factor in this chapter and the common emitter current 
amplification factor will be exclusively h lt . 



Amplifiers with Negative Feedback 439 

fact, G 34 is usually a voltage dividing network, as will be seen in the ex- 
amples which follow. Notice that the reference gain is reduced by the 
factor 1 + KzaP. 

Negative feedback may be used to stabilize the gain of an amplifier. 
With reference to Eq. 1 1.6, the product G 23 G 3 4 may be large in comparison 
with unity. Under these conditions the transfer function with feedback 
becomes 

G 23 _ J_ 

Gad 



"13 — 



G23G34 



(11.8) 



Similarly, the reference gain of the amplifier becomes 



K v 



1 



(11.9) 



Again, G 34 may be the ratio of two resistances in a voltage-dividing net- 
work, and thus the transfer function, or gain, is almost wholly independent 
of the characteristics of the amplifying device which may change with 



*i 



O 



*2 



*7 



*3 



Xi 






*5 



Gs6 



G 6 7 ■* 



*6 



Fig. 11.2. A feedback amplifier with two input signals. 



temperature, age, and power supply voltage. For example, the trans- 
conductance of a vacuum tube decreases with age. However, the gain of 
the amplifier will remain essentially constant as long as G^G^ (or K 23 fi) 
remains large in comparison with unity. 

More than one signal may be introduced into an amplifier, as is shown 
in Fig. 1 1.2. This feedback amplifier has two input signals, x 1 and x t . The 
plus signs indicate that x 3 is added to x t ; however, these two signals may 
have different waveforms or frequencies and the instantaneous polarity 
would then vary from time to time. 

If the amplifiers are linear, the law of superposition may be applied 
to the signals in the amplifier of Fig. 1 1.2. That is, the part of the output 
signal « 6 which results from input signal x 1 may be determined as though 
signal x 4 were not present. Likewise, the part of the output signal x e 
which results from input signal x t may be determined by considering x 1 to 



440 Electronic Engineering 

be nonexistent. The sum of these two output signals would actually be x e . 
Equation 11.6 may be used to obtain the transfer functions with feedback. 



Also 



1 ■+■ C7 2 3"56«67 

G ie = - -£« (11.11) 

1 + G^GszGm 

With no feedback (using again the principle of superposition), G 16 = 
G 23 G 56 an d G ie = G m . Therefore, it is evident by inspection of Eq. 1 1.10 
and 11.11 that the negative feedback reduces the transfer function from 
each input by the same amount; namely, the total loop gain of the system 
plus one. 

Now, consider the case where the signal x t is distortion 3 or noise 
generated within the amplifier. The application of negative feedback would 
reduce the amount of the undesirable signal appearing in the output. The 
desired signal from x x would also be reduced by the same amount. How- 
ever, it may be possible to increase the desired signal sufficiently to com- 
pensate for the reduction in gain caused by the feedback. Thus the ratio 
of the undesired signal to the desired signal could be reduced by the same 
amount as the reduction of gain in the amplifier. The success of this plan 
hinges on the possibility of obtaining additional distortion-free, or noise- 
free amplification ahead of the undesired disturbance. This additional 
distortion-free amplification is no problem in the case of distortion because 
the nonlinearity of an amplifier increases with signal level. Therefore, the 
final or output stage of an amplifier is usually the chief contributor of 
distortion. Conversely, the reduction of noise by feedback is not so fruitful 
because the noise in the output of an amplifier is determined primarily by 
the noise generated in the first stage. For example, the noise generated in 
the input of the first amplifier will be amplified and then applied as a 
signal to the second amplifier, as discussed in Chapter 9. Therefore, 
the most noise-free amplifier should be used as the first, or input, amplifier. 
It is usually assumed that the signal input to the first amplifier is fixed. 
Thus the application of negative feedback will reduce both the signal 
and noise by the same amount and the signal to noise ratio will remain 
unchanged. 

3 The consideration that the signal x t might represent the amplifier distortion appears 
to be in conflict with the requirement that the amplifiers must be linear to validate the 
superposition theorem. However, the representation of the distortion as an external 
signal effectively removes this distortion source from the amplifier, thus creating a 
distortionless amplifier. 



Amplifiers with Negative Feedback 441 

Returning to the reduction of distortion by negative feedback, we see 
that the ratio of distortion with feedback D f , to distortion without 
feedback £>„, is 

Df GJ(l + GjaGggGg?) _ 1 (11.12) 

D G 56 1 + Gzfiufiw 

But G 23 G 56 is the transfer function from the desired signal input to the 
output or total forward transfer function. Therefore, in the symbolism of 
Fig. 11.1, 

Ih = 1 (11.13) 

D 1 + G 23 G 3 4 

In the region of operation where the reference gain is applicable, Eq. 11.13 
reduces to 

D. ^— (1114) 

1 + **,/? 

11.2 THE EFFECT OF NEGATIVE FEEDBACK ON BANDWIDTH 

The effect of negative feedback on the transfer function of a single 
stage R-C coupled amplifier will be investigated first. The transfer function 

which neglected the shunt capacitance was 

6 p., 2d 2 -"■> 

G=-K—?— ° (7.28) 

The negative sign associated with AT will be omitted in the remainder of this 
chapter. This sign indicates polarity reversal which has already been 
accounted for in the derivation of the feedback formula. Then, if it is 
assumed that the feedback is obtained from a resistive voltage divider so 
that G 34 = /S, the transfer function with feedback is 

Ksl(s + co 1 ) = Ks 

Ks P s + co 1 + Ksp 

S + «>! 

r _ Ks = ^ ^ (11.16) 

' (1 + Kp)s + o>! 1 + K0 s + oVO + K/S) 

It may be observed from Eq. 1 1.16 that w,, as well as the reference gain, is 
reduced by the factor 1 + A"/S. Since the transfer function of the trans- 
former-coupled amplifier is of the same form as that of the R-C coupled 
amplifier when the shunt capacitances are neglected, the reduction in (o t 
would be the same in the two cases. 



442 



Electronic Engineering 



The effect of feedback on the large 5 or high-frequency performance of 
an R-C coupled amplifier will now be investigated. The transfer function 
of the R-C coupled amplifier without feedback follows: 



G = - 



Kco 2 
s + co 2 



(7.50) 



Again, it will be assumed that negative feedback is obtained from a resistive 
voltage dividing network so that G 31 = /? is constant and real. Then 



G f = 



G,= 



Kco 2 j(s + a> 2 ) 



Km, 



1 + Kficozlis + co 2 ) s + o> 2 -f K(ico 2 

K m 2 (l + Kp) 

1 + Kp s + to 2 (l + Kp) 



(11.17) 



(11.18) 



Notice in Eq. 11.18 that co 2 is multiplied by the factor 1 + Kp. Therefore, 
the upper cutoff frequency, and hence the bandwidth, of the amplifier is 
increased by the same factor by which the reference gain is reduced. 




(i+W 



Fig. 11.3. The frequency response of a typical R-C coupled amplifier. Curve A is for 
the amplifier with no feedback. Curve B is taken with negative feedback but no increase 
of input gain. Curve C is for negative feedback and increased input signal. 



Consequently the gain-bandwidth product of the amplifier with feedback is 
the same as that of the amplifier without feedback. " " — 

The effect which negative feedback has on the frequency response of an 
R-C coupled amplifier is illustrated in Fig. 11.3. The improvement of 
the frequency response characteristics is most evident when the input signal 
is increased to compensate for the decreased gain due to feedback, as 
shown by Curve C. The rise time and flat top response to a rectangular 
input signal are improved by the same factor as the frequency response, 



Amplifiers with Negative Feedback 44S 

when feedback is applied. The relationships between frequency response 
and time response were discussed in Chapter 7. 

The effect of negative feedback on the high-frequency performance of 
a transformer-coupled amplifier will be deferred to Section 11.7 in which 
the stability and performance characteristics are considered jointly. 

In passing, the student should observe that the equations which have 
been developed for negative feedback also apply for the positive feedback 
if the sign of j8 is reversed. Then the factor 1 + AT/J becomes 1 - Kfi and 
the performance of the amplifier is modified in a manner opposite to that 
of negative feedback. 

PROB. 11.1. A certain amplifier has a reference gain = 100. Negative feed- 
back is applied to the amplifier. If the feedback factor /? = 0.1, determine the 
reference gain of the amplifier with feedback. 

PROB. 11.2. The amplifier of Prob. 1 delivers 10 w of power output at 10% 
distortion when 0.1 v signal is applied to the input without feedback. Determine 
the per cent distortion and required input voltage after the negative feedback has 
been applied, assuming that the power output is to remain at 10 w. Answer: 
D f = 0.91% and V { = 1.1 volts. 

PROB 11.3. A frequency response curve of the amplifier of Prob. 11.1, without 
feedback, indicates that/i = 30 Hz and/ 2 = 100 KHz. The amplifier has R-C 
coupling. Determine/i and/ 2 after the application of negative feedback. Sketch 
the frequency response with feedback and without feedback. 

The equations in the preceding paragraphs of this chapter are completely 
general in regard to the types of signal and transfer function involved. For 
example, the transfer function could be a current ratio (current gain), a 
voltage ratio (voltage gain), or a ratio of current to voltage. However, the 
only transfer functions which have been considered for electronic ampli- 
fiers are current gain and voltage gain. Where current gain is being 
considered, the feedback factor must be a current ratio. Conversely, when 
the voltage gain is of prime concern, the feedback factor must be considered 
as a voltage ratio. Every amplifier has a voltage transfer function as well 
as a current transfer function. The relationship between these transfer 
functions is 

G = Ys = h£± = £& (11.19) 

where G v = the voltage transfer function 
G t = the current transfer function 
Z L = the 5-domain load impedance 
Z; = the ^-domain input impedance 
Similarly, every feedback network must provide a ratio of feedback 
current to'output current as well as a ratio of feedback voltage to output 



444 



Electronic Engineering 

voltage. These ratios will generally not be the same, as is demonstrated 
from the distortion relationship, Eq. 11.14. Consider first that the 
transfer functions are voltage ratios ; then 



D,= 



D n 



(11.20) 



1 + KJ V 

Substituting the reference current gain for the reference voltage «ain 
(Eq. 11.19), we find that ° 

Df = ^ = — ±- (1121) 

But the distortion in the output current waveform is the same as the dis- 
tortion in the output voltage waveform when the load is resistive. Also, a 
specific feedback network will provide a specific amount of distortion 
without regard to whether the total loop gain has been considered as a 
current ratio or a voltage ratio. Then, from Eq. 11.21, 



A*, = v K t 



and 



K, 



ft = 



R, 



(11.22) 



(11.23) 



11.3 THE EFFECT OF NEGATIVE FEEDBACK ON OUTPUT 
IMPEDANCE 

Figure 11.4 will be used to investigate the effect of negative feedback 
on output impedance. The output resistance is determined by applying a 









K 


«A 


^_ 


'2 


t 




R ° il'° 


f 

"2 

{ 1 










7— 1 












































Fig. 11.4. A circuit used to determine the output impedance of an amplifier with voltage 
feedback. 



445 



Amplifiers with Negative Feedback 

voltage v 2 to the output terminals and by evaluating the current i a which 
results. The impedance will be determined in the mid-frequency range 
and will, therefore, be resistive. Note that the output resistance R of the 
amplifier without feedback is drawn external to the block representing 
the amplifier. Since the feedback voltage /U , is proportion al to the outpu t 
™u afri this t ype of feedback (parallel output connection) is known as 
voltage feedback . The input current is 

,• = /?hEs (11.24) 

Since the amplifier must include an odd number of polarity reversals 
(negative feedback), the output current i A is in the direction shown. 

t A = Kiii = &f± (11-25) 

If the feedback network draws negligible current in comparison with the 
output current, the total current provided by the source v 2 is 

h=io+iA = ^ + ifi Vi (11.26) 

But, using Eq. 11.23, PJR { = &/*„• Then 

'*-(i + f) (1, ' 27) 

and since K£ t = Kjt, = Kft, the output resistance with feedback is 

p = g» = R ° (11.28) 

Kof i 2 1 + Kfl 



In the second case to be considered, the feedback voltage v, is pro- 
portio nal to the output current as shown in Fig. 11.5. This type of feedback 
i Tknown as curr e nt feedback. The amplifier input current is 

U=-Pii, < n - 29 ) 

Then the amplifier output current is 

i A = Kt i t = -|W. (H-30) 

The current / 2 as provided by the source v 2 is 

i, = /. + '4 = '.-^A (11 - 31) 

If the voltage drop i 2 R f across the series feedback resistor is small in 



446 



comparison with v 2 , then i ~ v 2 lR and 



R„ 



Electronic Engineering 



(11.32) 



As a result, the output resistance with feedback is 



R - v *. 

K of — — ' 



«.a + w 



(11.33) 



Thus the output impedance has been increased by the term (1 + Kfi) as 
a result of current feedback. 

Observe from Fig. 11.5 that the polarity of the amplifier input voltage 







i 


K 


«A 


»2 


o- r 




R B < 


> |*o 


s 








v 2 














P ^Rf 






^-<8,i2 


c 


t'2 















Fig. 11.5. A circuit used to determine the effect of current feedback on output impedance. 

resulting from feedback is reversed in the current feedback case as 
compared with the voltage feedback of Fig. 11.4. Note, however, that 
the feedback is negative in each case, as can be verified by terminating the 
amplifier with a load resistance and by comparing the polarity of the 
feedback voltage with an assumed input voltage. 

PROB. 11.4. The cathode follower may be considered as an amplifier with feed- 
back since v gk = v t - u . In this case, /?„ = 1. Derive expressions for the voltage 
gain and output impedance of the amplifier in terms of n, r p , and R L . 
PROB. 11.5. An amplifier with impedance in the emitter circuit may be treated 
as an amplifier with feedback. The feedback voltage is i L Z E and /?„ = Z E jZ L 
where Z E is the emitter impedance, Z L is the load resistance, and i L is the 
output current. Determine the voltage gain of a vacuum tube amplifier which 
has fi = 100, r p = 50 KH, R L = 100 Kil, and R E (unbypassed resistance in the 
cathode circuit) = 2 Kfi. Answer: G f = 28.6. 

PROB. 11.6. Determine the output resistance of the amplifier of Prob. 11.5. 
Would you expect the bandwidth of this amplifier to be greater with the cathode 
resistance bypassed or unbypassed ? Why ? 



Amplifiers with Negative Feedback 



447 



11.4 THE EFFECT OF FEEDBACK ON INPUT IMPEDANCE 

In the preceding discussion, there has been no concern about the 
method of mixing the feedback signal with the input signal. There are 
two basic ways of mixing these signals— the amplifier input and feedback 
ports may be connected either in series or parallel. The series connectio n 




Fig. 11.6. A circuit used to determine the input resistance with the amplifier input and 
feedback ports in series. 

of Fig. 11.6 will be considered first and the input impedance for this 
configuration will be determined. The input voltage with feedback is 

»„ = »< + /U, = »«(1 + *X) ( n - 34 > 

But v t = URi. Then, (using Kj v = Kp) 

«,„ = i^l + */S) (11-35) 

Finally, the input resistance with feedback is 



Rif = V J1 = Ri (l + KP) 



(11.36) 



The inquisitive reader will ask, "What effect does this increased imped- 
ance have on the gain of the amplifier which precedes the feedback 
amplifier?" The answer is, "If the increased impedance provides an 
improved impedance match between the two amplifiers, the power gain of 
the preceding amplifier will be increased and vice versa." For a vacuum 
tube amplifier, the input impedance of the tube at mid-frequencies is 
essentially infinite; therefore, the impedance match remains essentially 
unchanged and the reference gain of the preceding amplifier is unchanged. 
In a transistor amplifier, the input impedance without feedback is usually 



448 



Electronic Engineering 



smaller than the output impedance of the driving amplifier. Under these 
conditions, the increased input impedance of the feedback amplifier will 
increase the power gain of the preceding amplifier for moderate amounts 
of feedback because the impedance match will be improved. As the input 
impedance of the transistor amplifier is increased, its input current 
will decrease and the current gain of the preceding amplifier will 
decrease accordingly. However, this is the actual mechanism by which 
the current gain of the feedback amplifier is reduced, and it has been 
accounted for by the previously derived transfer functions. The voltage 



'if 


>- 


>_ 


K 




lo 

>- 




r 










.,! 












































Fig. 11.7. A circuit used to determine the input resistance with the amplifier input and 
feedback ports in parallel. 

gain of the preceding amplifier will increase more rapidly than its current 
gain decreases until the matched condition is obtained. This explanation 
accounts for the improved power gain. 

The parallel port connection is shown in Fig. 11.7. From this con- 
figuration, the input current with feedback is 

i if = ii + i f (11.37) 

But /, = fijig and /'„ = /Q/,. Therefore, 

i if = /, + K#& = 1,(1 + KA) (11 -38) 

Since /', = v ( Y { , where Y f is the amplifier admittance without feedback, 
the input admittance with feedback is 



Y it = hl =Y i {\ + Kf3) 



(11.39) 



Observe that for the parallel connection the input impedance is reduced 
by the factor 1 + AT/?. 



Amplifiers with Negative Feedback 449 

Again, the effects of the reduced input impedance of the feedback 
amplifier on the gain of the preceding amplifier should be considered. In 
a transistor amplifier the current gain of the preceding amplifier is essen- 
tially unaltered, assuming the normally mis-matched R-C coupled case. 
Voltage and power gain will decrease, however. In a vacuum tube ampli- 
fier, the power gain of the driving amplifier will be increased for moderate 
amounts of feedback because of the improved impedance match. The 
feedback equations account for the reduction on voltage gain. 

The feedback paths previously discussed have all been provided by 
resistors. It was assumed that the output should be a faithful reproduction 
of the input. Sometimes it is desirable to alter the waveforms or frequency 
response of the output as compared with the input. This alteration can be 
accomplished by including capacitive or inductive elements in the feedback 
path. Fnr pyampltv a series capacitance in the feedback circuit will provid e 
a feedback factor of the form 

G 34 = i?-^— (11-40) 

The feedba ck factor will then be essentially inversely proportional to s for 
values" ofTsmaller than co f . where to, is the reciprocal of the time constant 
of the feedback circuit. This type of circuit could be used to provide a 
" bass boost" tone control.. 

Another interesting use of the feedback circuit to control the waveform 
of the output is the integrator version of the operational amplifier which is 
used so extensively in analog computers. The parallel-type feedback is 
used but the feedback path consists of a capacitor. This path causes the 
input impedance of the amplifier to be of the form Y { = \jCs. Under 
these conditions the output waveform will be the integral of the input 
waveform. 

PROB. 11.7. Show that the feedback factor G 34 can be written as Py^r^ when 
the feedback path consists of R and C in series. ' 

PROB. 11.8. A given transistor amplifier has 2K.H input resistance and 2KQ 
output resistance without feedback. The output voltage polarity is opposite 
to that of the input. What type of input and output port connections (series or 
parallel) are needed for a negative feedback circuit which will permit this ampli- 
fier to be inserted in a 500 n (characteristic impedance) telephone line without 
causing a mis-match in the line in either direction? (i.e., make Z in = Z ou t = 
500 O). What value of Kp will be required? Answer: K0 = 3. 

11.5 TYPICAL FEEDBACK CIRCUITS f^44< 

Since low output impedance is usually desirable in an amplifier— at 
least a broadband amplifier — voltage feedback is much more common than 
current feedback. Figure 11.8 shows a common series voltage feedback 



450 



Electronic Engineering 



arrangement for an R-C coupled amplifier. Since there is no polarity 
reversal between the emitter and collector of an amplifier, the second 
amplifier provides the only polarity reversal between the point of applica- 
tion of the feedback and the amplifier output. This polarity reversal is 
necessary to provide negative feedback. Notice that the feedback voltage 




Fig. 11.8. Series-type voltage feedback for R-C coupled amplifiers. 



tends to cancel the input voltage. It is usually desirable that R x be as small 
as possible to provide minimum degeneration and thus maximum gain for 
the input amplifier. Also, the resistance of R 2 should be large in compari- 
son with the total shunt output impedance -of the output amplifier. Other- 
wise R 2 would appreciably reduce the gain of the second amplifier due to 
loading. Consequently, a sensible design procedure might be to choose 
R 2 about ten times the total shunt resistance R sh of the output amplifier. 
Then the resistance of J? x could be calculated to give the desired amount of 
feedback. 



Amplifiers with Negative Feedback 451 

A casual glance at Fig. 1 1.8 might cause us to conclude that the reference 
feedback factor is 

a = l Al = Rl — (11.41) 

(,(«! + R 2 ) R x + R 2 

However, the emitter current of the input amplifier in addition to the 
feedback current i, flows through R t . Moreover, these two currents are 
interdependent, which adds to the complexity of calculating an accurate 
value of Ri. An easy solution to the problem is to calculate R x from the 
relationship of Eq. 11.41 and then experimentally adjust R 2 after the 
amplifier has been constructed. An example of this procedure follows. 

Example 11.1. For the purposes of stabilizing the transfer function, or gain, and 
improving the response of the tube amplifier of Fig. 1 1.8, it is desired to make 
the reference loop gain AT/3 = 10. First, the reference gain K must be calculated 
without feedback. We will assume that the reference voltage gain is found to be 
50 for each stage. Then the total reference gain of the amplifier is 2500. Conse- 
quently |8 = 10/2500 = 0.004. 

The shunt resistance of the output stage would be somewhat less than r„ — 
perhaps 50 KO since high /« tubes were used. Then the resistor R 2 in the feedback 
circuit should be of the order of 500 Kii. Using 470 KQ for R 2 and the value of /J 
above in Eq. 1 1.41, « x is calculated to be about 1.9 KO. This value is a typical 
value of cathode bias resistance for a triode amplifier. Therefore, the additional 
bypassed resistance shown in the figure will not be needed. Also a resistance of 
this magnitude would cause appreciable degeneration in the input amplifier. 

The desired loop gain could be adjusted experimentally by measuring the 
open loop reference gain of the amplifier with the resistor R 2 disconnected. Then 
a value of R 2 could be selected experimentally which would reduce the gain of the 
amplifier by the factor 1 + Kfi = 1 + 10 = 11. 

In case there is no blocking capacitor in the feedback loop, as shown in Fig. 
1 1 .8, the direct current through R 2 must be added to the quiescent plate current of 
the input amplifier in determining the bias for that amplifier. 

PROB. 11.9. Determine a feedback circuit for the transistor amplifier of Fig. 
1 1.86. Use 2N192 transistors and assume that the load resistance of the second 
transistor is 2 KQ. Is your value of R t suitable for temperature stabilization of 
the input transistor? Use A"/S = 10. 

A parallel voltage feedback arrangement for R-C coupled amplifiers is 
shown in Fig. 11.9. In this arrangement the feedback loop must include 
an odd number of stages to provide negative feedback. This statement is 
true only if each amplifier is a common emitter configuration. More 
correctly, an odd number of polarity reversals must be included within 
the feedback loop to provide negative feedback. Figure 11. 9a shows three 
common emitter amplifiers within the loop whereas Fig. 11.96 includes 



452 



Electronic Engineering 




(a) 




(b) 
Fig. 11.9. Parallel voltage feedback applied to R-C coupled amplifiers. 



only one common emitter stage within the loop. Since no blocking capaci- 
tor is included in the feedback path, there is d-c feedback as well as signal 
feedback. In Fig. 11. 9a this d-c feedback tends to stabilize the quiescent 
operating point of the input transistor. For example, an investigation of 
the relative polarities of the amplifiers with parallel feedback, such as those 
of Fig. 11.9, will show that the d-c feedback tends to stabilize the Q point 
of the input amplifier as well as the final amplifier. Thus, feedback can 
improve the thermal stability of a transistor amplifier. However, in order 
for the d-c feedback to affect any stage other than the first in the feedback 
loop, the amplifier must be direct-coupled. In the series feedback arrange- 
ments of Fig. 11.8, the d-c feedback decreases the thermal stability of the 
input amplifier. Therefore, a blocking capacitor would probably be 



Amplifiers with Negative Feedback 453 

included in the feedback path, to prevent this degeneration in thermal 
stability. 

In a transformer coupled amplifier, negative feedback can improve 
linearity and response characteristics but not to the same degree as that 
shown for the R-C coupled amplifier. The extent of the improvement will 
be seen more clearly in the following section on stability. Transformers are 
seldom used as coupling devices in untuned amplifiers except to couple the 
final amplifier to the load. However, the feedback loop should include 
the transformer so that the effective transformer characteristics will be 
improved by the feedback. A typical feedback loop which includes the 
output transformer is shown in Fig. 11.10. When a transformer is included, 
negative feedback may be obtained with an arbitrary number of stages of 
arbitrary connection within the loop because either end of the transformer 
secondary may be used as the reference, or grounded, end. Again, R 2 
should be large in comparison with R L , but /?! should be small in order to 
minimize the degeneration in the first stage. Parallel feedback could have 
been used just as well as the series type shown. In the parallel circuit R t 
may serve as the biasing resistor R\, providing that it is a suitable value. 




Fig. 11.10. A feedback circuit which includes the output transformer. 



454 Electronic Engineering 

As the foregoing discussion indicates, many variations and applications of 
feedback circuits may be devised by the resourceful engineer. 

11.6 STABILITY OF FEEDBACK CIRCUITS 

In the beginning of this chapter it was shown that the transfer function 
of an amplifier with negative feedback is 

Gi3 = r^V (1L6) 

where G^ is the forward transfer function of the amplifier and G 3i is the 
s-domain feedback factor. The magnitude of the transfer function can be 
infinite (poles can exist) if there are values of s for which the denominator 
of Eq. 1 1 .6 becomes zero. Then 

1+G 23 G 34 = (11.42) 

or 

G 23 G 3i = - \/Q = 1 /180° (11.43) 

If Eq. 11.43 can be satisfied at real frequencies {s = y'w), a pole will lie on 
they'co axis of the s plane and any transient initiated in the amplifier will not 
decay. Thus the amplifier can have a steady-state output when there is no 
input. This condition is known as oscillation and the amplifier is then 
said to be unstable. Oscillation is a "must" for an oscillator, but it is 
intolerable in an amplifier. 

The instability, or oscillation, may occur in a feedback amplifier 
because of the inductances and capacitances in the circuit which may cause 
polarity reversal with reference to the mid-frequency (s) range of the 
amplifier. We can determine from the overall transfer function of the 
amplifier, Eq. 1 1 .6, whether the amplifier will be stable or unstable. One 
way to determine the stability would be to replace s in the transfer function 
by jco and then to test the denominator and find if it can become zero at real 
frequencies. A much more informative method is to plot on the s plane 
all the values of s for which Eq. 11.42 and 11.43 hold. If any part of this 
plot crosses the ju> axis, the amplifier will be unstable for values of loop 
gain which exceed a certain, determinable amount. This plot is known as 
a root-locus plot. In addition to determining whether a particular amplifier 
is stable or not, this root-locus technique gives an insight into the design 
of a feedback amplifier which not only will be stable, but which will 
provide the desired response characteristics. 

The root-locus plot is actually a plot of the locus of all the values of s 
which will give the proper angle (180° for negative feedback) of G^G 3i 
(see Eq. 11.43). We may gain the impression that tedious work would be 



Amplifiers with Negative Feedback 455 

required in plotting a root locus. On the contrary, simple rules have been 
developed which make it possible to sketch a sufficiently accurate locus 
with very little effort. Some of these rules are given in Appendix II. The 
derivation or formulation of all of these rules is too lengthy to be included 
in this work, but may. be found in textbooks and literature primarily 
devoted to this technique. 4 A few of the basic rules will be stated in the 
paragraphs which follow. 

A few examples will illustrate how the root-locus plot may be used to 
determine the stability and response characteristics of an amplifier. A 
single R-C coupled amplifier with a purely resistive feedback circuit as 
shown in Fig. 11.9 will first be considered. The transfer function derived 
in Chapter 7 for the R-C coupled amplifier follows: 

G *£^ (11.44) 

(s + ft^Xs + <w 2 ) 
Then 

Ksa> 2 



_ (s + wjjs + w 2 ) (11.45) 

13 Kfaco 2 



1 + 



(s + fc>i)(s -1- ft> 2 ) 



Poles occur in the overall transfer function, G 13 , at values of s which 
reduce the denominator of Eq. 11.45 to zero. This denominator will 
equal zero when the following equality holds. 

K P sm * = -1 = l /l80° (11.46) 

(s + eo^s + co 2 ) 

This equation is of the form 

K'F{s) = 1 /180° (11-47) 

where K' is the constant part of G^G^, in this case Kpm* and the function 
F(s) is the part of G^G^ which varies with J. Then 

f(s) = ! (11.48) 

(s + ft)j)(s + w.) 

A root-locus plot of F(s) will now be made. First, the poles and zeros 
of F(s) will be plotted on the s plane (see Fig. 11.11). Inspection of Eq. 
1 1 .48 shows that the angle of F(s) can be 1 80° when s is real and is less than 
zero but greater than -co,. Then the numerator is negative and the 

1 An excellent treatment is given in Introduction to Feedback Systems, L. Dale Harris, 
Wiley, New York, 1961. 



456 



Electronic Engineering 



denominator is positive. The same situation occurs when s is real and less 
than — « 2 . In fact, the general rule given in Appendix II states that 
portions of the real axis lying to the left of an odd number of singularities 
are branches of the root locus in a negative feedback circuit. These plots 
are shown in Fig. 11.11. Another rule in Appendix II states that the 
number of infinite seeking branches is equal to the difference between 
the number of finite poles and finite zeros. Or, to invoke still another rule 
from Appendix II, all branches must begin on a pole and end on a zero. 
The number of infinite seeking branches in the example is one since there 
are two poles and one zero. The plot shown in Fig. 11.11 is the complete 



JO) 



— 0>2 



-0>! 



Fig. 11.11. The root-locus plot of an R-C coupled amplifier. 



root locus because there is one infinite seeking branch on the negative real 
axis. Notice that the number of locus branches is the same as the order of 
the denominator of F(s). 

As previously mentioned, poles will occur in the overall transfer 
function when the angle of K'F(s) is 180°, which limits s to the root-locus 
branches, and when the magnitude of K'F(s) is one. Then 



K' 



1 



IWI 



(11.49) 



If both of the foregoing conditions can be satisfied by a value of s which 
has a real part a which is either zero or positive, the transient part of the 
output either remains constant or increases with time, respectively. Then 
the amplifier is unstable. Conversely, when the poles of the overall transfer 
function all lie in the left half of the s plane, the transients decay exponen- 
tially with time and the amplifier is said to be stable. In case the root-locus 
branches lie entirely within the left half plane, the amplifier is said to be 
unconditionally stable, which means it is stable for all finite values of gain. 

It may not be obvious that the R-C coupled feedback amplifier whose 
root locus is plotted in Fig. 11.11 is unconditionally stable, since the root 



Amplifiers with Negative Feedback 



457 



locus touches the jw axis at the origin. But there is a zero of F{s) at the 
origin; therefore A' must be infinite at the origin and the amplifier is stable 
for any finite gain. Thus this circuit is unconditionally stable. 

It is also well to notice that the magnitude of F{s) is infinite at a pole of 
F(s). Therefore, from Eq. 1 1.49, A" must be zero at a pole of F(s). Conse- 
quently, the value of A" and hence the magnitude of the loop gain must 
increase from zero to infinity as a root-locus branch is traversed from a 
pole to a zero. 




Double zero 



Fig. 11.12. Root-locus plot for a two-stage R-C coupled feedback amplifier. 



The case of two R-C coupled stages included in the feedback loop will 
now be investigated. In this case, 

K 1 K 2 Pu) il u)2 2 s 



CaAs = K'F(s) = 



(s + co u )(s + w 12 )(s + co 21 )(s + <o 22 ) 



(11.50) 



where w n is m 1 of the first amplifier, etc. A pole-zero plot of F(s) is made 
in Fig. 11.12. The root locus is also sketched as shown with the aid of the 
basic rules irf Appendix II. This locus is tangent to thejco axis at the origin, 
but the point of tangency occurs at a double zero, so A" must be infinite at 
this point and the amplifier is unconditionally stable. 

Each branch of the root locus [beginning on a pole of F(s) and ending 
on a zero] contains all values of A' between zero and infinity, as previously 
mentioned. If a point which represents the same value of A' is found on 
each branch, these points are the poles of the overall transfer function G 1S 
with the loop gain adjusted to give the value of A'. This condition follows 
from the fact that these points are all the zeros of the denominator of G 1S 
at the value of A'. Thus it may be seen from Fig. 11.12 that a pair of 



458 Electronic Engineering 

complex conjugate poles may appear in the transfer function of the two- 
stage R-C coupled feedback amplifier. These poles cause the circuit to 
have a damped oscillatory transient response. As the total loop gain 
increases, the damping ratio £ decreases since a remains essentially 
constant andjco increases with increasing K'. Thus the value of loop gain 
which provides optimum transient response can be found. For example 
a damping factor £ = 0.8 is near optimum for some applications. To 
determine the value of K' which will provide a predetermined value of £, 
the constant £ line is drawn on the pole zero plot as shown in Fig. 11.12, 
where £ = cos 6. This constant £ line intersects the root-locus branch as 
shown at point a in Fig. 11.12. The values of $ at this point a is substituted 
into the equation F(s). Then K' is the magnitude of l/F(s). A device 
known as a spirule can also be used to determine K' at a given 
value of s. 

A feedback amplifier which has three R-C coupled stages included 
within the loop will now be considered. In this case 

/- r vet \ ^ K 2 K 3 ftco 21 co 22 ft) 23 s 

«2 3 34 = A. t (S) = 

(s + u> n )(s + w 12 )(s + u> 13 )(s + a> 21 )(s + w 22 )(s + co 23 ) 

(11.51) 

The pole-zero plot and root-locus sketch for this configuration is given in 
Fig. 11.13. Two branches of the locus cross the positive ju> axis at points 
a and b. This configuration is known as conditionally stable, which means 
that the feedback amplifier is stable for some values of loop gain but 
unstable for other values. The amplifier here will be stable for all values 

of K' less than the smaller of K' a or K' b , where K\ 
1 

W) 

than K\ the frequency of oscillation will be w a , and vice versa. The values 
of K' a and K\ could be evaluated either by substituting jw„ and ja> b 
respectively into l/F(s) or by using the spirule. 

Although Fig. 11.13 shows all the significant poles and zeros of the open 
loop transfer function, it is usually not practical to include both the poles 
near the origin and those far removed from the origin on the same plot. 
This follows from the fact that poles which result from the shunt capaci- 
tance occur at values of s which may be hundreds of times as large as the 
values of s at poles near the origin. Thus a scale which would permit the 
inclusion of the poles far removed from the origin would cause the poles 
and zeros near the origin to be compressed into an area about the size of a 
pinpoint. One solution to this problem might be the use of logarithmic 



F(a) 
, and F(a) is the function F(s) evaluated at point a, etc. If K' a is less 



and K\ = 



Amplifiers with Negative Feedback 



459 



scales for the s-plane plot, but then the root-locus rules as well as the 
spirule would be unusable. Fortunately, the singularities far removed 
from the origin have essentially no influence on the root-locus branches 
near the origin and vice versa, so the problem can be divided into two 
parts, as shown in Fig. 11.14. This solution is basically the same as the 




Fig. 11.13. Root-locus plot of a three-stage R-C coupled amplifier with feedback. 



technique used in Chapter 7 whereby one equivalent circuit was used to 
represent the low-frequency range and another equivalent circuit was used 
to represent the high-frequency range. 

Although the amplifier will be stable if the loop gain is adjusted so that 
all the poles of the overall transfer function G 13 are in the left-half plane, 
the transient response of the amplifier may be unsatisfactory, as mentioned 
before, if the poles are too near the jco axis. Again, the desired damping 
ratio may be obtained by adjusting the loop gain to give the value of K' 
which is found at the intersection of the root locus with the desired 
damping ratio line. Points c and </(Fig. 11.13) are both intersections with 
the I = 0.8 line. The value of K' should be determined at both points, 



460 



Electronic Engineering 




3 poles far 

removed 

from the 

origin 



A. 3 poles and 
^■3 zeros near 
the origin 




Fig. 11.14. The root-locus plot of a three-stage R-C coupled amplifier divided into two 
regions, (a) Singularities far from the origin; (b) singularities near the origin. 

and, if very little overshoot can be tolerated, the lower value of loop gain 
should be used. 

From the foregoing examples, a general trend should be noticed. 
That is, the larger the number of stages included within the feedback loop, 
the more limited the loop gain must be in order to maintain good transient 
response and stability. 

PROB. 11.10. A three-stage R-C coupled tube amplifier has feedback from out- 
put to input. All amplifiers are identical and have a*! = 100 and co 2 =0.5 x 10 6 
radians/sec. Using a root-locus plot similar to the one in Fig. 11.12, determine 
the largest value of loop gain Kfi which can be used without causing high- 
frequency oscillation in the amplifier. What would be the frequency of oscillation 
if oscillation does occur? Answer: Kfi < 8, eo = 8.66 x l(fi radjsec. 
PROB. 11.11. An R-C coupled transistor amplifier has two identical stages 
which have / 2 = 100 KHz, and / x = 20 Hz. Determine the loop gain which 
will give f = 0.8 when feedback is applied around the amplifier. 



11.7 THE CHARACTERISTICS AND STABILITY OF 

TRANSFORMER COUPLED FEEDBACK AMPLIFIERS 

The stability of a feedback amplifier which consists of one transformer 
coupled stage will now be considered. The approximate transfer function 
of a transformer coupled amplifier was developed in Chapter 7 and is 
repeated as follows: 

— (11.52) 



G = 



(s + W!)(s + a + ja> 2 )(s + a — jco 2 )(s + a> 3 ) 



Amplifiers with Negative Feedback 4*1 

Then, from the poles and zeros of the open loop transfer function (Eq. 
1 1.52) the root locus is drawn in Fig. 11.15. Observe that the single-stage 
transformer-coupled feedback amplifier is only conditionally stable. As 
the loop gain K' is increased, all the poles of the overall transfer function 
move away from the poles of the open loop transfer function shown in 
Fig. 11.15. The complex pair move in the direction of increasing cw and 
decreasing a. This action increases the high-frequency response but 
decreases the damping ratio £• As the pole which leaves co x moves toward 
the origin, the low-frequency response improves. In addition, the effect 




1 pole and 
1 zero near 
the origin 



Fig. 11.15. The root-locus plot for a single-stage transformer-coupled amplifier. 

of the pole moving from w 3 toward — oo is to increase the magnitude of 
the transfer function at large values of s, or in other words, to improve the 
high-frequency response. In summary, then, the frequency characteristics 
of the amplifier are improved but the transient response may be deterio- 
rated. 

A paradox exists in the application of feedback to a transformer 
coupled amplifier: the transformer with poor frequency and transient 
response is less susceptible to improvement by feedback than is the 
high-quality transformer. Figure 11.15 shows that the damping ratio £ 
generally decreases as the loop gain is increased. Therefore, the transient 
response of the amplifier deteriorates as the feedback is increased unless 
the damping ratio of the transformer itself is less than the optimum value. 
High-quality transformers usually have a comparatively high damping 
ratio. When negative feedback is applied to amplifiers using these 



462 Electronic Engineering 

transformers, the transient response, frequency response, and distortion are 
improved simultaneously by moderate amounts of feedback. However, 
when negative feedback is applied to the amplifier which incorporates a 
poor transformer with a damping ratio of 0.8 or less, the frequency response 
can be modestly improved at the expense of impaired transient response. 
Therefore, those amplifiers which are in greatest need of improvement are 
generally the least susceptible to improvement by feedback. 

The root-locus plot for a feedback amplifier consisting of two typical 
transformer-coupled tube amplifier stages is shown in Fig. 11.16. Notice 




2 poles and 

2 zeros near 

the origin 



Fig. 11.16. The root-locus of a two-stage transformer-coupled feedback amplifier. 



that a modest loop gain would improve the frequency response with little, 
if any, deterioration of the transient response. However, as K' is increased, 
they'w axis is approached quite rapidly and poor transient response if not 
instability would result if a moderately large loop gain were employed. The 
magnitude of loop gain which could be tolerated would naturally depend 
on the characteristics of the transformers used. It is evident from Fig. 
11.16 that the loop gain could be increased if the poles of the transformers 
were moved to the left. This could be accomplished by reducing the 
leakage inductance and distributed capacitances of the transformers. 
Also, the transient response of the amplifier would be improved if the 
transformer damping ratios were increased. 

The sketch of the root locus for a typical transformer-coupled amplifier 
plus an R-C coupled amplifier is given in Fig. 11.17. The design of a 
transformer-coupled feedback amplifier is more difficult than the design of 
an R-C coupled feedback amplifier because the constants of the transformer 



Amplifiers with Negative Feedback 



463 



are not usually known. Some experimentation will no doubt be necessary 
in optimizing the design. One satisfactory procedure might be to determine 
the desired loop gain in view of the distortion or other response require- 
ments of the amplifier, then estimate the number and type of stages which 
might be included within the feedback loop without risking poor transient 
response or instability. A test of the amplifier after it has been constructed 
will either verify the assumptions or indicate the necessity for modifications. 
More than one stage should usually be included in the feedback loop for 
two reasons. First, the open loop gain should be large enough so that the 




2 poles and 

2 zeros near 

the origin 



Fig. 11.17. The root-locus plot for a transformer-coupled amplifier and an R-C coupled 
amplifier. 

feedback factor /? may be small and the feedback network will not load the 
output amplifier. Second, the signal input to the feedback amplifier 
should be small so that the preceding amplifier will be a small-signal, 
low-distortion amplifier. Except for these two requirements, the applica- 
tion of negative feedback to each individual amplifier stage has some real 
advantages when compared with the inclusion of several stages within one 
feedback loop. 5 

PROB. 11.12. Sketch the root locus of a feedback amplifier which includes one 
typical transformer-coupled and two typical R-C coupled amplifiers. 



11.8 COMPENSATING NETWORKS 

The impression may have been gained while studying the stability of 
feedback amplifiers that some additional zeros would be invaluable for the 

5 An excellent discussion of single versus multiple feedback systems is given in 
Introduction to Feedback Systems, L. Dale Harris, Wiley, New York, 1961. 



4*4 Electronic Engineering 

purpose of improving the stability of a feedback amplifier. Although 
zeros, by themselves, are difficult to buy at any price, a rather simple 
compensating network which provides an additional pole along with a 
zero will materially increase the loop gain which may profitably be used in 
a multistage feedback amplifier. This network is commonly known as a 
phase lead network and is shown in Fig. 11.18. The transfer function of 
the feedback network is 



a,-Z- 



R* 



G,= 



G f = 



V R 2 + (R 1 lsC)l(R 1 + 1/sC) 

R* _MsCR 1 + 1) 



R 2 + Rj(sCR 1 + 1) R 2 (sCR 1 + I) + R 1 
1 



s + 



KjC 



S + <Og 



s + 1/R X C + 1/R 2 C s + 



(11.53) 
(11.54) 

(11.55) 



v f 



•R 2 



Ri 

-vVWVV- 



Fig. 11.18. A phase lead network. 



where w„ = 1/Z^C and m b = l/Z^C + l/R 2 C. Since R t is normally large 
in comparison with R 2 , a>„ may be large in comparison with «„. In fact, 
since the feedback factor /? is the value of G, when j is small in comparison 
with both co a and w 4 , Eq. 11.55 may be used to obtain the relationship 
w 6 = w a/£- The manner in which this network may be used to advantage 
is illustrated in Figs. 11.19 and 11.20. In this case, a feedback amplifier 
including a transformer and two R-C coupled stages is considered. In 
Fig. 11.19 the root-locus plot is made for the amplifier with a purely 
resistive feedback network. Figure 11.20 is the root-locus plot of the 
same amplifier with phase lead compensation in the feedback network. 
Notice how the effect of the pole at w 3 has been essentially neutralized by 
the zero of the compensating network at co a . The pole of the compensating 
network at co b is far to the left, and thus moves the intersection of the 
root-locus asymptotes quite a distance to the left compared with the point 
of intersection in the uncompensated case. Consequently, the root-locus 



Amplifiers with Negative Feedback 



465 




3 poles and 

3 zeros near 

the origin 



Fig. 11.19. Root-locus plot for a feedback amplifier consisting of two R-C coupled 
stages, one transformer-coupled stage, and resistive feedback. 

branches which leave the complex poles depart in a more nearly vertical 
direction and traverse a much greater distance before crossing the ja> axis 
than they did in the uncompensated feedback amplifier. Therefore the 
amplifier would be stable at larger values of loop gain, or conversely, 
for a given value of loop gain the amplifier would have better transient 
characteristics when phase lead compensation is used in the feedback 
network. 

Trial and error calculations will show that the ratio of &>„ to w in the 
compensating network is more important than is the actual value of m a or 
co,,. A practical upper limit of m b and co a results from the requirement that 




Fig. 11.20. The root-locus plot of the amplifier of Fig. 11.19 with phase lead 
compensation. 



466 



Electronic Engineering 




250 v - 
Fig. 11.21. A feedback amplifier with phase lead compensation. 

the shunting capacitor be large in comparison with the stray capacitance 
in the feedback network. 

The circuit diagram of a feedback amplifier with phase lead compensa- 
tion is given in Fig. 1 1.21. The directly coupled phase inverter eliminates 
one pole-zero pair near the origin. This system is used in the popular 
"Williamson" amplifier. 




24-26 v 
Control or 

filament 
transformer 



Fig. 11.22. A quasi-complementary symmetry transistor amplifier employing feedback. 



Amplifiers with Negative Feedback 467 

The circuit diagram of a transistor feedback amplifier is given in Fig. 
11.22. This circuit is basically the quasi-complementary symmetry circuit 
discussed in Chapter 10 (see Fig. 10.41). The 2N2712 transistor has been 
added to the previous circuit to allow the addition of negative feedback 
without increasing the input voltage requirement. The type feedback 
used also provides very high input resistance. Therefore, this amplifier 
is a desired load for a wide variety of sources such as radio tuners, phono- 
graph and microphone pre-amplifiers, etc. An additional feature has been 
incorporated in this circuit. The filter capacitance in the power supply is 
provided by two equal capacitances, so that the mid-point of the power 
supply may be used as a signal reference or ground. This feature is 
desirable because the entire circuit appears as a bridge-type circuit with 
the power supply attached to one pair of the bridge terminals and the 
signal applied to the other pair. Therefore, the amount of power supply 
ripple voltage coupled into the desired signal is very small. Thus the hum 
level in this amplifier is very low even though the power supply filter is not 
very good. An additional bonus of this balanced power supply arrange- 
ment is the reduction in the number of capacitors required. Observe that 
the filter capacitor C 2 also bypasses the emitter circuit resistor R t in the 
2N2712 circuit and the two filter capacitors block the direct current from 
the load resistor. 

Note in designing the circuit of Fig. 11.22 that the emitter resistor R E 
is chosen to give the desired voltage gain in the 2N2712 stage, since 
G„ ~ RJRe- Then the feedback resistor R, is chosen to give the desired 
amount of feedback. Observe that the d-c feedback factor is much larger 
than the a-c feedback factor because R t is included in the d-c circuit but 
not in the a-c circuit, and i^ is large in comparison with R E . This large 
amount of d-c feedback provides very good Q-point stability for all the 
amplifiers in the circuit. 

11.9 FEEDBACK DESIGN FROM GAIN-PHASE (BODE) PLOTS 

The gain and phase plots, known as Bode plots, which were discussed in 
Section 9.3, can be conveniently used in the design of feedback amplifiers. 
Any feedback amplifier will become unstable if the feedback signal is 
in-phase with the input signal and has a magnitude equal to (or greater 
than) the input signal. Consequently, even a negative feedback circuit 
can be unstable (as the reader has already been shown) if the open-loop 
gain is equal to (or greater than) one and a total phase shift of 180° occurs 
in the open circuit loop. Since the Bode plots contain both gain and phase 
information, these plots can be used in the design of feedback circuits as 
shown in the following example. 



468 



Electronic Engineering 



Example 11.1 A three-stage R-C coupled amplifier has three zeros at s = 0; 
three poles at s =j2n(l0); and single poles at s =j2t<10 4 ), * =j2n(3 x 10 4 )', 
and s =j2M9 x 10 4 ). The total reference gain K = 1000. Investigate the 
conditions under which this circuit will be stable if it is used in a negative feedback 
configuration. 

The power gain and phase plots of this amplifier are shown in Fig. 11.23. 
The amplifier will be unstable if the total loop gain magnitude is at least unity 
(Odb) at a frequency which produces 180° phase shift, relative to the mid- 



60 



50 



40 



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20 



10 









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120 




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V 






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1 
















1 








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1( 

ue 


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ncy h 


6 


1 


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e 
3 


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6 


— 1 — lOU 

10 6 



Fig. 11.23. Plots of power gain and phase (relative to mid-frequency phase) for the 
amplifier in Example 11.1. 

frequency phase. Note that in this example there are two such frequencies, 
fx and f y . (Compare these frequencies with /„ and /„, respectively, in the root- 
locus plot of Fig. 11.13 which represents a somewhat similar situation.) The 
gain magnitude off x is 44 db and at/„ is 39 db. Therefore, the feedback circuit, 
if purely resistive, must provide a gain less than -44 db or the amplifier will be 
unstable. In fact, the feedback gain must be considerably less than -44 db 
or poor transient response will result because the j-plane poles are too near 
theyco axis. Most designers allow an extra 6 db to 10 db, known asgain margin 
in order to provide good transient response. In this example a 6-db gain margin 
will be considered adequate and the desired feedback factor is 

20 log p = -50 db 

P = log- 1 (-2.5) =0.00316 

Kfi = 3.16 

This value of permissible loop gain is disappointingly low. However, the 
feedback factor and loop gain could be increased if at least one of the low 
frequency poles were eliminated by direct coupling. Then the gain at/;, would 



Amplifiers with Negative Feedback 469 

be the limiting gain (39 db) and the desired feedback factor would be 

/? = log- 1 (-2.25) =0.0056 

and Kp = 5.6. This improvement results from the staggering of the upper 
cutoff frequencies, or poles, as compared with the triple pole at the low frequency. 
The permissible loop gain could again be greatly increased if a phase-lead 
compensating network were included to shift/, to a much higher frequency. 

PROB. 11.13 One stage of the amplifier in Example 1 1.1 is made direct coupled 
so one of the poles at s =J2-*{10) is eliminated. In addition a phase-lead 
network (as shown in Fig. 11.18) is used as the feedback path with R t = 10 KU, 
R 2 = 200 fi, and C = 1100 pf. 

(a) Make an open-loop gain-phase plot for this circuit. 

(6) Determine the value of Kp. 

(c) Find the frequency at which the relative phase shift is 180 . 

(d) Find the gain margin for the closed-loop circuit. Answer: (c) / <* 
3 x ICPHz (d) Gain margin ~ 16 db. 

PROB 11.14. The distortion of the amplifier of Fig. 11.22 without feedback 
is approximately 6% at maximum power output. Use the values you deter- 
mined for the circuit of Prob. 10.28 and determine the additional components 
needed for the circuit of Fig. 11.22 to provide 0.6% distortion and 1-v peak 
input at maximum power output. 

PROB. 11.15. The amplifier of Fig. 11.21 uses 12AX7 tubes for the voltage 
amplifiers and 6CM6 tubes for the power amplifier. The load resistance is a 16 12 
speaker. The transformer has complex poles at ±j<o = 2n x 20,000 radians/sec, 
£ = 9 and a real pole at « 3 = -2* x 40,000 radians/sec. The upper cutoff 
frequency of the R-C coupled stage is 100 KHz. With the aid of the tube maniud, 
calculate the approximate open loop gain, then calculate the value of R, which 
will reduce the distortion in the output to one-tenth of the value without feedback. 
Make a root-locus plot of the feedback amplifier, without the phase lead capacitor 
and determine whether the amplifier will be stable at the desired value of total 
loop gain. If so, what will be the damping ratio £? (Refer to the locus branches 
from the complex pair.) 

PROB. 11.16. Using a phase lead feedback network for the amplifier of Prob. 
1115 determine the value of C which will essentially cancel the pole of the trans- 
former which lies on the real axis (o> 3 ). Make a root-locus plot of the amplifier 
and determine the maximum value of total loop gain which can be used without 
causing oscillation. Determine the damping ratio £ (caused by the complex poles 
of the transformer) for the value of total loop gain required to reduce the dis- 
tortion by a factor of ten. 
PROB. 11.17. A single R-C coupled amplifier has a transfer function 

' jKco, 

C™ = 



(s + (ojis + <o 2 ) 



(a) If a*! is 100 and co ? is 10 6 radians/sec, sketch a root-locus plot for a three- 
stage R-C coupled amplifier with negative feedback. 

(b) If singularities far from the origin are ignored, what value of K is required 
for oscillation and what is the frequency of oscillation? 



470 



Electronic Engineering 




Output 



Fig. 11.24. The circuit to be used for Prob. 11.18. 



PROB. 11.18. A circuit is connected as shown in Fig. 11.24. Find the gain of 
this circuit and determine the magnitude of the feedback parameter p. 
PROB. 11.19. A feedback circuit contains one transformer-coupled amplifier 
and an R-C amplifier. If the transfer characteristics of these stages are as given 
below, sketch the root-locus plot. For what values of loop gain will the circuit 
be stable? 



R-C Stage G = 



sK.COa 



Trans stage G = 



(s + (Oj)(s + 0) 2 ) 

sKa> 3 co 5 2 



+ °>tKs + a +j(o s Xs + a -jco s )(s + co 3 ) 



co 2 = 10 6 
COj = 10 2 

co 4 = 2 X 10 2 
a = 10 6 
co s = 10 6 
a> 3 = 2 X 10 6 
tf=500 



PROB. 11.20 Make a Bode plot of the amplifier of Problem 1 1.19 and determine 
the value of /S if the gain margin is 6 db. 



12 



Large-Signal Tuned 
Amplifiers 



In Chapter 10, the class B amplifier was found to have a higher theoretical 
efficiency than the class A amplifier. However, unless special circuits were 
used, the distortion was very high. In general, this distortion takes the 
form of adding harmonics to the desirable fundamental. Consequently, if 
a method can be found to eliminate the harmonics, a very desirable circuit 
is achieved. For broad frequency band amplification, the class B push-pull 
circuit achieves both high efficiency and even harmonic cancellation. 
However, if a narrow band of frequencies is to be amplified, a different 
approach is possible. A resonant circuit is used to pass the desired band of 
frequencies and to reject the harmonics. 

In class B and class C amplifiers, the collector (or plate) current flows 
for only a portion of each cycle. Consequently, the collector current can 
be considered as a series of pulses with a pulse repetition rate equal to the 
frequency of the input signal. An insight into the behavior of this type of 
signal can be gained by considering the action of a linear tuned amplifier 
which is activated by a square wave. 

471 



472 



Electronic Engineering 



The response of the tuned amplifier (Fig. 12.1) to a periodic pulse can be 
found by using the concepts already developed in Chapter 8. In Section 
8.1 the response of a tuned amplifier stage to a step function (Fig. 8.4) was 
found to be 



„ = _ Is£ e -w sin mot 



(8.18) 



Now, if the input signal is composed of a square wave with a period of 
l// > (Fig- 12.2a), this input signal can be synthesized from a series of 




(a) (b) 

Fig. 12.1. A class B or C tuned amplifier, (a) Tube circuit; (6) transistor circuit. 

positive and negative step functions as shown in Fig. 12.2c. The output 
signals (as given by Eq. 8.18) for the step functions are plotted in Fig. 
\2.2d. When these individual output signals are added together, the total 
output signal (Fig. 12.26) results. Hence, sinusoidal output signals may 
be obtained when the input signals are square waves. 

One way to visualize the behavior of the tuned amplifier is to reduce the 
square wave of Fig. 12.2a to a series of sinusoidal voltages. Fourier 
analysis allows the square wave to be reduced to a fundamental component 
and an infinite series of harmonic components. In the tuned amplifier, 
the tuned circuit in the output of the amplifier is tuned to the same fre- 
quency as the fundamental component in the square wave. Hence, the 
impedance of the tuned circuit is very high for the fundamental component. 
Since the gain of the amplifier is proportional to the plate or collector load 
impedance, the gain of the amplifier is high for the fundamental compo- 
nent. In contrast, the capacitor C (Fig. 12.1) has a low impedance for all 
harmonic components. Consequently, the gain of the amplifier at these 
harmonic frequencies is very low. 

From the foregoing analysis, the tuned amplifier is able to amplify the 
fundamental component of a class B or class C amplifier while suppressing 



Large-Signal Tuned Amplifiers 



473 





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474 Electronic Engineering 

the harmonics which are generated. Since the theoretical efficiency of a 
class B amplifier is higher than the theoretical efficiency of a class A 
amplifier, the reader may suspect that the theoretical efficiency of a class C 
amplifier may be even higher. Such is the case, as will be demonstrated 
later in this chapter. As a result of this higher efficiency, the usual 
large-signal tuned amplifier is a class C amplifier. 

Since the usual large-signal tuned amplifiers are class C, the remainder 
of this chapter will be directed to the design and analysis of class C 
amplifiers. Of course, the ideas pertaining to class C amplifiers can be 
extended to include class AB or class B amplifiers, so the methods outlined 
are rather general. 

PROB. 12.1. A tube is connected as shown in Fig. 12.1. Assume g m is 2000 
micromhos and r v is 1 meg fl. The capacitor C has a value of 400 pf and L has an 
inductance of 10 mh. If the Q of the tuned circuit is 50, (a) find the voltage gain 
of this amplifier for a signal with the same frequency as the w of the tuned 
circuit ; (b) find the gain of this amplifier for a signal with a frequency of 2co . 
Answer: (a) G v = -400, (b) G v = 6.67&S1. 



12.1 BASIC BEHAVIOR OF CLASS C TUNED CIRCUITS 

The class C amplifier is biased below cutoff. Hence, if a sinusoidal input 
waveform is applied, the plate or collector current flows for less than 180° 
of a cycle. The actual angle over which current flows is called the conduc- 
tion angle. Since the plate or collector circuit contains a fairly high Q- 
tuned circuit, the voltage variation in the plate or collector circuit is 
essentially a sinusoid. Of course this plate or collector voltage is 180° 
out of phase with the input voltage when the output circuit is tuned to the 
input signal frequency. Accordingly, the voltages and currents in a typical 
class C tube amplifier are as shown in Fig. 12.3. In this amplifier, the 
control grid is driven positive. In fact, the control grid is usually driven 
positive to such an extent that the maximum control grid potential is equal 
to the minimum plate potential. This condition usually is used to obtain 
maximum power output. If the control grid becomes more positive than 
the plate, the control grid may draw so much current that the plate 
current actually decreases. Hence, the optimum power output usually 
occurs when the maximum control grid potential is equal to the minimum 
plate potential. 

The class C amplifier requires a graphical analysis that is different from 
the methods employed previously in this text. The methods used in the 
preceding chapters assumed that the plate current was a fairly linear 
function of the input voltage. However, in the class C amplifier this linear 
relationship no longer holds. In fact, as will be shown in Section 12.2, the 



Large-Signal Tuned Amplifiers 



475 




Fig. 12.3. Voltage and current relationships in a class C amplifier. 






300 



200 



V f = 7.5 volts 
Grid -No. 2 volts = 500 
Grid -No. 2 amperes = I G2 
Grid-No. 1 amperes = J G1 
Plate amperes = I P 




-100 



-200 



7000 



3000 4000 

Plate volts 
Fig. 12.4. Constant current curves for a 4-1000A tube. (Courtesy of Radio Corporation 
of America.) 



476 Electronic Engineering 

plate current may actually have a dip when the grid voltage is maximum 
positive. However, Fig. 12.3 shows that the relationship between grid 
voltage and plate voltage is linear when the plate circuit is tuned. Conse- 
quently, a very useful set of tube characteristics 1 is one in which the plate 
voltage is plotted as a function of grid voltage with the plate current held 
constant. Such a set of characteristic curves is shown in Fig. 12.4. These 
curves are known as constant current curves. The constant current curves 
can be plotted from the information contained in a set of conventional 
plate characteristic curves. Notice that the tube amplification factor is the 
reciprocal of the slope of the constant current curve. 

PROB. 12.2. Draw a set of constant current curves for a 6J5 tube. The plate 
characteristic curves are given in Appendix III. 

12.2 GRAPHICAL CLASS C SOLUTION FOR VACUUM TUBES 

The previous section indicated that class C amplifier problems can be 
solved by using a different graphical approach. In this approach, the 
constant current curves are used. In many cases, these constant current 
curves are furnished by the tube manufacturer. 

The usual method of approach to a graphical solution for vacuum tube 
class C tuned amplifiers is as follows : 

1. Plot the quiescent operating point on the constant current curves. (See point 
Gof Fig. 12.6.) 

2. Determine the desired magnitudes of the maximum grid voltage and the 
minimum plate voltage. As these voltages occur at the same time this point 
is also plotted on the constant current curves. (See point P of Fig. 12.6.) 

3. Since the a-c plate voltage is 1 80° out of phase with the control grid excitation, 
the plot of instantaneous plate potential vs the instantaneous grid potential 
is a straight line. 2 Thus, a straight line can be drawn on the constant current 
characteristics connecting the two points found in step 1 and step 2. (See the 
line between point P and point Q of Fig. 12.6.) 

4. Values of plate current and grid current (also screen current for tetrodes and 
pentodes) can be found for given values of grid voltage. Thus, the waveform 
for all voltages and currents can be plotted as shown in Fig. 12.3. 

5. The average values of these currents can be found by integrating the current 
waveforms over one complete cycle. In addition, a Fourier analysis can be 

1 The analysis of the transistor class C amplifier is more complex than the analysis of 
a tube amplifier. Consequently, the tube amplifier will be considered first and then the 
transistor amplifier will be treated. 

8 If the phase angle between two a-c voltages of the same frequency is different than 0° 
or 180°, a plot of one voltage vs the other voltage is an ellipse. In fact, the dimensions 
of the ellipse can be used as a means of determining the phase angle between the two 
voltages. For a more detailed analysis of this approach see F. E. Terman, Radio 
Engineers Handbook, McGraw-Hill, New York, pp. 947-949, 1943. 



Large-Signal Tuned Amplifiers 477 

used to determine the fundamental components of these current waveforms. 
6. When the average currents are known, the power requirements from the power 
supplies can be determined. In addition, with the fundamental current com- 
ponents known, the signal power input and the signal power output can be 
determined. 

In order to find the average (d-c) currents and the fundamental com- 
ponents of the signal currents, integration must be performed. Usually, a 
graphical approach is necessary. Therefore, a review of graphical integra- 
tion may be in order at this time. To integrate the area under the curve of 
Fig. 12.5, the base of the area is divided into a series of uniform lengths 




Fig. 12.5. A method of graphical integration. 

Ar as shown. Now, since the integral of y from a to b is the total area 
under the curve or the sum of all the small areas, 

J>=*[(? + ?) + (? + !) + (? + ?) + "] (121) 



or 



j"y dx ~ A*[| + 2/ 2 + 2/s + 2/4 + • • • + 2/n-i + ^] (12-2) 

Of course, the smaller A» becomes, the more accurate the graphical 
integration will be. 

The method of solution for a class C amplifier can be further clarified 
by the use of an example. 

Example 12.1. A 4-1000 A tube is to be used as a class C amplifier in a circuit 
similar to Fig. 12.1. The screen grid is maintained at a constant potential of 
500 v. The characteristic curves of this tube are shown in Fig. 12.4. If a 5000-v 
plate supply and a -200-v bias supply is to be used, find the characteristics of 
the amplifier. 



478 



Electronic Engineering 



o 

Z 



o 




-100 



-200 



2000 



3000 4000 

Plate volts 



5000 



6000 



7000 



Fig. 12.6. Graphical solution of a class C amplifier. (Curves are Courtesy of Radio 
Corporation of America.) 



TABLE 12.1 
A List of Current and Voltage Values in a Class C Amplifier 

Phase Grid 1 Plate Plate Grid 2 Grid 1 

Angle 6 cos Voltage Voltage Current Current Current 



0° 


1.000 


+ 150v 


150 v 


2.2 a 


1.2 a 


0.3 a 


10° 


0.985 


+ 144.5 


250 


2.4 


1.1 


0.26 


20° 


0.940 


+ 129 


475 


2.4 


0.8 


0.18 


30° 


0.866 


+ 103 


800 


2.2 


0.45 


0.093 


40" 


0.766 


+68 


1275 


1.85 


0.18 


0.048 


50° 


0.643 


+25 


1900 


1.2 


0.029 


0.005 


60° 


0.500 


-25 


2550 


0.47 


0.005 





70° 


0.342 


-80.4 


3300 


0.03 








80° 


0.174 


-139.1 


4150 











90° 


0.000 


-200 


5000 












Large-Signal Tuned Amplifiers 47 " 

The constant current curves are reproduced in Fig. 12.6. The quiescent oper- 
ating point is denoted in Fig. 12.6 as the point Q. The second point is taken 
where the control grid potential is maximum and the plate potential is minimum. 
As already mentioned, the usual optimum point occurs when the maximum grid 
potential is approximately equal to the minimum plate potential. Let the control 
grid swing to +150 v and let the plate swing down to +150 v. This value of 
grid and plate voltage is shown in Fig. 12.6 as point P. The line from P to Q of 
Fig. 12.6 is drawn and is known as the operating line and determines not only 
the relationship between the control grid voltage and the plate voltage, but also 
the relationship between the control grid potential and the various currents in the 
tube. Accordingly, the line PQ is marked for grid voltage intervals corre- 
sponding to At intervals. In this example, At is chosen so that coAt is equal 
to 10 electrical degrees. 

From the operating line in Fig. 12.6, the values of voltages and currents in 
Table 12.1 are found. The values listed in Table 12.1 can be used to plot the 
various voltages and currents in the circuits. This plot is given in Fig. 12.7. 
Also from the values in Table 12.1, the magnitudes of the d-c currents and the 
fundamental components of the grid and plate signal currents can be found. 

We can simplify the graphical Fourier analysis by choosing the proper value 
as reference. For example, in Fig. 12.6, if t = or = at v„ max, only cosine 
terms will be present in the Fourier series because the value of the function at 
+ 0° will be equal to the value of the function at -0°. Because of this simplifi- 
cation, the v„ max point of Fig. 12.7 will be used as the reference (0 = 0) in the 
following analysis. The 10° intervals along the operating line can be located by 
first determining the grid or plate voltage at these intervals. 3 

Now, from the Fourier analysis of Fig. 12.7, the d-c value of plate current I P 
is given by 



-7- [" ipdO 02.3) 



In Fig. 12.7, the value of A0 is 10° or tt/18 radians. Hence, the graphical integra- 
tion from to n becomes 

£ ijpM = L ^f +ip2+ip3 + ...+ i„ + *f ) (12.4) 

where i P1 , ipi, etc. have the values indicated in Fig. 12.7. Now, since the area 
under the curve (Fig. 12.7) from -it to is equal to the area under the curve 
from to it (for the given reference), the average plate current is (Eq. 12.3) 

For the values in Table 12.1, the value of I P is 



/ 



~ 1 (— + 2.4 + 2.4 + 2.2 + 1.85 + 1.2 + 0.47 + 0.03 + - ) 
18 \ 2 2 / 



• For example, at these intervals, v 01 = -200 + 350 cos in A0) where A0 = 10° 
and n = 0, 1, 2, 3, etc. The values of v 01 at these intervals are listed in Table 12.1. 



480 



Electronic Engineering 



+9850 



>+5000 



+ 150 



+ 150 



-200 



-550- 




Time- 



Fig. 12.7. Voltages and currents of the class C amplifier in Example 12.1. 



11.65 



= 0.647 amp 



By similar analysis, the direct or average current to the screen grid and to the 
control grid can be found. For the values in Table 12.1, the magnitude of I ai 
(the average control grid current) is 44.5 ma and the magnitude of I G2 (the 
average screen grid current) is 175 ma.. 
The peak value of the fundamental component of the plate current is also 



Large-Signal Tuned Amplifiers 481 

needed. To find this value, the required Fourier expansion is 

hi = ' I >p cos e de (12 - 6) 

Because of the symmetry involved, this can be written as 

2 r* 
4i = - »'p cos e de < 12 - 7 > 

"Jo 
Accordingly, 

I vl ~ - x -^ r ' PlC ,° S °° + i> 2 cos 10° + ips cos 20° + i Pi cos 30° 

+ ■■■ + J P(n -i) cos (lOn - 20)° + i Pn cos (lO/i - 10)° (12.8) 

For the values in Table 12.1, the peak magnitude of the fundamental value of 

plate current, I vl , is 

10.059 
I P1 ^— - 9 1.118 amp 

By similar analysis, the peak magnitude of the fundamental component of 
control grid current I gl , is 81.7 ma. The magnitude of the fundamental com- 
ponent of screen current could also be found, but the screen has essentially no 
a-c component of voltage. Consequently, the power at the fundamental fre- 
quency is zero, and the magnitude of fundamental current is not required. 

From the values of voltages and currents just derived, the power relationships 
of the tube can be established. The power drawn from the power supply by the 
plate of the tube is 

i» 7 = V PP I P (12.9) 

For this example Pj = 5000 x 0.647 = 3235 w. 

The load impedance which the tuned plate circuit must offer at the funda- 
mental frequency can now be found. Since the tuned circuit is at resonance, 
this impedance is purely resistive and has a value given by the relationship 

where V v and I Pl are the peak values of fundamental voltage and current. In 
this case, 

4850 

^ = uT8= 434on 

Now the signal power output P is 

Vylj,\ (^PP ~ E'gmliOJgl (12 11) 



482 Electronic Engineering 

For this example, 

(5000-150)1.118 
Pg = 1 _? = 2710 w 

The total power dissipated by the plate of the tube is 

P d =P I -P (12.12) 

Thus, P d = 3235 - 2710 = 525 w. This value of plate dissipation is well within 
the limits of the tube, so operation at this level is permissible. 
The plate efficiency ri v of the tube is given by the relationship 

^=JxlOO (12.13) 

For this example, 

2710 
"» = 3235 =83 - 8% 

Other information which can be found is the effective input resistance. This 
resistance Rm is given by the relationship 

R lB = ^i (12.14) 

Thus, 

350 

^=ao8T7= 4280 " 

where V gl and I gl are the peak values of the control grid voltage and the funda- 
mental component of grid signal current. The grid driving power P g can be 
found by the relationship, 

P g =^ (12.15) 

In this example, 

0.0817 x 350 
P g = - = 14.3 w 

Since the maximum control grid dissipation for a 4-1000 A tube is 25 w, this grid 
power dissipation is permissible. However, the stage which supplies the signal 
for this stage must be capable of furnishing the grid power required by this stage. 
The results of this example are tabulated below. 

The peak signal voltage on the grid V gl is 350 v. 

The peak fundamental component of grid current I gl is 81.7 ma. 

The input resistance R ia is 4280 CI. 

The average d-c voltage in the grid circuit V ao is —200 v. 

The average d-c current in the grid circuit I G1 is 44.5 ma. 



Large-Signal Tuned Amplifiers 483 

The input power to the grid circuit P g is 14.3 w. 

The average d-c current in the screen circuit I aa is 175 ma. 

The voltage on the screen grid V G2 is +500 v. 

The average d-c current in the plate circuit I P is 0.647 amp. 

The average d-c voltage on the plate V PP is 5000 v. 

The peak signal voltage on the plate V v is 4850 v. 

The peak fundamental component of plate current I pl is 1.118 amp. 

The required value of plate load impedance R L is 4340 £2. 

The power output P is 2710 w. 

The plate power input P/ is 3235 w. 

The plate dissipation P d is 525 w. 

The plate efficiency »?„ is 83.8%. 

The plate conduction angle is 17° to (180° - 17°) or 146°. 

The foregoing example illustrates the method of approach to solve class 
C (or class B or class AS) problems. The list of known quantities at the 
end of the example indicates the effectiveness of the method. Unfortu- 
nately, if one bias or signal voltage level is changed, the entire process must 
be repeated for the new condition. Consequently, in order to optimize a 
circuit a number of solutions must be tried. Hence, a trial and error 
approach is indicated. On the brighter side, a little experience in the design 
of class C amplifiers reduces the trial and error to a minimum. 

The waveform of the plate current in Fig. 12.7 deserves a word of 
explanation. At maximum grid voltage the control and screen grids are 
robbing the plate current from the plate. Since the control grid is so much 
nearer the cathode than the plate, this action is not unexpected. In 
addition, since the screen grid is so much more positive than the plate, the 
screen has a tendency to take a larger portion of the total space current. 
An examination of the constant current curves (Fig. 12.6) indicates the 
reason for this type of behavior. Over most of the range of the constant 
current curves, these curves are straight. However, at the end of the curves 
where the plate voltage is very low, the curves rise gradually at first and 
then more sharply. As the control grid potential becomes more positive, 
the plate potential for the beginning of the rise becomes more positive. 
If the plate does not swing low enough to shift the operating line into the 
rising portion, the plate current will have a waveform indicated by Fig. 
12.8, curve a. If the plate swings low enough, the operating line of Fig. 
12.6 may become parallel to the constant current curves over the low-plate 
voltage end of the operating line. If this condition occurs, the plate 
current is saturated for low-plate voltages. This condition is shown in 
Fig. 12.8, curve b. If the plate swings to an even lower potential, the 
operating line may actually recross one or more constant current curves. 
In this case, the plate current has the waveform shown in Fig. 12.8, curve c. 



484 



Electronic Engineering 



The curves of Fig. 12.7 were drawn to aid the reader in visualizing the 
operation of the amplifier and to illustrate the procedure of solution. 
However, in most solutions the sketch is not required. In fact, all the 
values required for the solution of the problem can be obtained from 
Table 12.1.* 




Fig. 12.8. Typical plate current waveforms in class C operation. In curve a, the 
minimum plate potential is much larger than the maximum grid potential. In curve b, 
the minimum plate potential is low enough to cause plate current saturation. In curve c, 
the maximum grid potential is high enough to rob the plate of current while the plate 
potential is minimum. 

PROB. 12.3. A 6J5 tube is to be opeiated class C. V GG is -30 v and V PP is 
+300 v. Assume the peak magnitude of control-grid signal is 34 v and the peak 
magnitude of the plate signal is 260 v. Plot the plate current waveform. Use the 
curves from Prob. 12.2. 

PROB. 12.4. Repeat Example 12.1 with V GQ = -200 v, V PP = 5000 v, and 
v g 2 — 50 ° v - However, in this problem let the maximum control-grid potential 
be 145 v and the minimum plate potential be 250 v. Compare the results of this 
problem with the results of Example 12. 



12.3 DETERMINATION OF THE TUNED-CIRCUIT PARAMETER^ 

In Example 12.1, many electrical quantities were found, but no attempt 
was made to determine the values of the tuned-circuit components except 
to note the magnitude of resistance the tuned circuit should offer at 
resonance. Actually, the configuration of the tuned circuit determines the 
method of approach that must be used. In general, all the ideas developed 
in Chapter 8 can be used for the large-signal tuned circuits. However, 

4 To aid in determining the values of Table 12.1, a plastic overlay is made by the 
Eitel-McCullough Tube Co. This overlay is known as the "No. 5 Tube Performance 
Computer" and is available from the Manager of Amateur Service Department, 
Eitel-McCullough, Inc., 301 Industrial Way, San Carlos, California. 



Large-Signal Tuned Amplifiers 



485 



there is one striking difference in the philosophy of design in large-signal 
as opposed to small-signal coupling circuits. In the small-signal amplifier, 
efficiency is not of prime concern. The power loss in the coupling circuit 
results in smaller stage gain than would be possible with lossless circuits, 
but the loss of a few decibels gain is not serious in a high-gain amplifier. 
As noted in Chapter 8, the bandwidth of the small-signal amplifier is of 
major importance because this type of amplifier is used in applications 
which normally require the selection of a specified band of frequencies and 
the rejection of all other frequencies. On the other hand, the large- 
signal tuned amplifier is usually used in radio transmitters or other 
applications where power output and waveform of the signal are important 




Fig. 12.9. A typical class C coupling circuit. 

but frequency selection is not. In the tuned power amplifier the desired 
output is a sinusoid or modulated sinusoid and the Q's of the coupling 
circuits need be only high enough to adequately eliminate the harmonics. 
The coupling efficiency increases as the circuit Q decreases, as will be 
shown. The efficiency is important in high-power amplifiers where 
increased efficiency results in materially reduced operating costs. 

Considering a single-tuned, inductively coupled circuit as shown in 
Fig. 12.9, the power loss in the tuned circuit is PR , where / is the current 
and R is the resistance of the primary coil at the resonant frequency. 
The power transferred to the secondary circuit is PR' L where R' L is the 
resistance coupled into the tuned circuit as a result of the load current 
flowing in the secondary. As discussed in Chapter 8, the ohmic resistance 
of the secondary is usually very small in comparison with the load 
resistance. Then the efficiency of the coupled circuit becomes 



V = 



coupled power 



loss power + coupled power 



x 100 = 



PR' 



I\R + R L ) 
R'l 



x 100 



R + R L 



x 100 (12.16) 



486 Electronic Engineering 

The efficiency may be obtained in terms of the coil Q and circuit Q, since 
R = a> LlQ and (R' L + R )=^- then 

R > L = ^k _ Rq = ^ _ «£ (12 . 17) 

L Q Q Qo 

= a> L(i/e - i/o 100 = o^e x 100 

co L(l/Q) Q 

It may be seen from Eq. 12.18 that Q must be large in comparison with 
the circuit Q in order for the efficiency to approach 100%. This relation- 
ship generally holds for coupled circuits. The maximum practical effi- 
ciency can be obtained by selecting a coil with a Q as high as possible and 
then choosing a circuit Q which is just adequate to provide good waveform. 
Values of circuit Q between 12 and 20 are commonly used. 

Example 12.2. The tube in Example 12.1 must operate at a frequency of 10 8 
radians/sec. The tuned circuit has an actual Q of 200 and the desired loaded 
Q is 20. The actual value of load resistance is 100 O. and the coupling circuit of 
Fig. 12.9 is selected to provide the power transfer. 

To simplify the calculations, we will assume that the resistance of coil L 2 is 
much less than R L and also that the reactance of Z, 2 at resonance is much less 
than R L . 

From Example 12.1, the resistance of the loaded tuned circuit at resonance 
is 4340 O. For a parallel tuned circuit, the Q is given by the relation 

*" m ' C (12.19) 



(o L x G 

where R v is the parallel resistance, G is the conductance of the resonant circuit, 
and C is the capacitance of the tuned circuit. Hence, for this example 

C=^= ™ =4.6xl0-«fds 
a> 4340 x 10 6 

The value of L t can be found from the relation 

1 



«>„ = 



(12.20) 



VL X C 
In this case 

^- 10"x4.6xlO- - 2 - 18xl<Hh 

The loaded tuned circuit parallel resistance R „ is 4340 £1. Part of this resistance 
R' L is coupled from the load back to the tuned circuit and the remainder of this 



Large-Signal Tuned Amplifiers 



487 



resistance R is due to the resistance in the tuned circuit. Since the unloaded 
Q is known (200), Eq. 12.19 can be used to calculate R . 



e =-7- = 200 = 



Rr. 



or 



ovLj " 10 8 x 2.18 x 10-* 

R = 200 x 2.18 x 10" 4 x 10 6 = 43,400 CI 
Now, the parallel combination of R and R' L must be equal to R v . 
Thus we can write 



R't. = 



R P R _ 4340 x 43,400 
R - R v ~ 43,400 - 4340 



= 4820Q 



From Eq. 8.86, the magnitude of L 2 can be determined if the coefficient of 
coupling k is known. If k is assumed to be 0.7, L 2 is 



U = 



k 2 R', 



(12.21) 



For this example 



2.18 x 10- 4 x 100 „„„ . 
U = —TTT-r. .c^r, = 9.22 //h 



' 2 ~ 0.49 x 4820 Ci 
From Eq. 12.18, the coupling efficiency is given as 
Q„-Q 



t) = 



200 20 

x 100 = 20Q x 100 = 90% 



Thus, in this example, 90% of the power output from the tube is coupled into 
the load and 10% of this power is lost in the coupling circuit. 

PROB. 12.5. Repeat Example 12.2 if the Q of coil L x is 100. How much power 
would be coupled to the load under these conditions? 

PROB. 12.6. The tube in Example 12.1 is connected to a coupling circuit which 
is shown in Fig. 12.10. If the resonant frequency is 1 MHz and the Q of the 

C c = 0.01 Mf 




Rgl 



Cziz 



Rsg 
HWW- 



R g 

20 k! 



T-Csj 



■R = 10 K 

; input to 
next stage 



Fig. 12.10. The circuit for Prob. 12.6. 



488 Electronic Engineering 

tuned circuit with load is 20, find the values of the tuned circuit components. The 
Q„ of coil L is 200 at resonance. 

12.4 CLASS C TRANSISTOR AMPLIFIERS 

When we arrived at the graphical solution of the class C amplifier by 
using constant current curves, we assumed that the input voltage and 
output voltage of the amplifier were both sinusoidal. This assumption is 
valid only if the effective output resistance of the stage which drives the 
amplifier is small in comparison with the lowest value of the input resistance 
of the amplifier. In either tube or transistor amplifiers this requirement 
may not always be met. In case the source resistance is not negligibly 
small, the input characteristics should be modified to include the effects of 
the source resistance, as was done in the determination of transistor 
distortion in Chapter 10. The effect of the source resistance will be 
included in the example of the transistor amplifier which follows. 

The constant current curves for the transistor can be developed from the 
collector and input characteristic curves. The collector and input charac- 
teristic curves of a transistor are given in Fig. 12.11. From these curves, 
the constant current curves of Fig. 12.12 can be drawn. The curves of 
constant base current are found directly from the v BE vs i B curves of Fig. 
12.1 1. When the constant base current curves have been drawn, the curves 
for constant collector current can be obtained from the i c vs v CE curves. 
The required values of i B and v CE are obtained from Fig. 12.1 1 and plotted 
in Fig. 12.12. 

The curves of Fig. 12.12 can be used if the impedance of the voltage 
source which excites the base of the transistor is negligible. However, if 
the impedance of the source is not negligible, a modification must be made. 
For example, if the circuit of Fig. 12.13 is to be used, the curves of Fig. 
12.12 must be modified to the form shown in Fig. 12.14. This modification 
is achieved by observing that (in Fig. 12.13) the open circuit generator 
voltage v s is 

v s = i B Z s + v BE (12.22) 

The curves of Fig. 12.14 are a plot of v s vs v CE rather than of v BE vs v CE 
as in Fig. 12.12. In the curves of Fig. 12.14 each constant base current, 
curve has been shifted upward by a voltage value of i B Z s , where Z s = 10 D. 
in this example. Since the collector current is a function of the base 
current, the constant collector current curves must be shifted upward to 
maintain their relative positions in relation to the new constant base 
current curves. With the curves of Fig. 12.14 available, the solution 
proceeds as shown in Example 12.1. Note, however, that a new set of 
curves must be drawn if the generator impedance is changed. 



Large-Signal Tuned Amplifiers 



489 



10 
9 


^350 


-400 n 
ma 


na 






1 1 

Stud temp = 


25°C 






-/-73OU ma 
///250 ma 


















8 


/, 


200 m 


a 
















7 
6 


/ 




















' 


^150 


ma 
















5 
4 




^12 


5 ma 




















100 m< 
-7 


i 














3 








50 ma 














2 










-25 rr 


1 
a 


5 ma 






1 




















10 




0.2 0.3 

I B , in amperes 



0.4 



0.5 



"0 0.1 

Fig. 12.11. The characteristic curves of a 2N1899 power transistor. 



490 



Electronic Engineering 




20 40 60 80 100 

Collector volts 

Fig. 12.12. Constant current curves for the transistor of Fig. 12.11. 

M 

ci £i3K£2 ^loo a 




Fig. 12.13. A Class C transistor amplifier. 



Large-Signal Tuned Amplifiers 

5.0 



4.0 



491 



3.0 



2.0 



1.0 



-1.0 









"i B = 400 ma 












\ 






\ 








3bO ma 










\ 






1 jj- 


\ \ 








-i B = 300 ma 










\\" 


.= 9a 






\\ 








_• 


= 250 ma 








v\ K 


ic = 8 


a 






\ \ 






_• 


= 200 ma 








\ 


v/^^'c™ 7 


a 




V 




1 




B = 150 ma 

1 








e- 




i c = 6 a 








i c _ 






1 
.i B — 100 ma 

1 














i c =4a 


^ 










*c 


= 3a 

Li B =5 


Oma 






v 












■ ic = 2 a 

= ip = 25 ma 


















i c = 


= la 
















i 


C=iB" 


= Oma 











































20 



40 60 

Collector volts 



80 



100 



Fig. 12.14. Modified constant current curves for the transistor when the signal voltage 
source has an internal impedance of 10 Ci. 



12.5 SELF-BIAS CIRCUITS FOR CLASS C AMPLIFIERS 

The use of two or three batteries (Fig. 12.1) or power supplies is un- 
desirable in class C amplifiers. Accordingly, means of self-bias for class C 
amplifiers was achieved at an early date. The circuits of Fig. 12.10 and 
Fig. 12.13 illustrate typical self-bias stages. The screen circuit of Fig. 12.10 
operates in the same manner as described in Chapter 6. As before, the 
value of R sg is given by the relationship 



R-sg — 



v PP - V. 



G2 



(6.13) 



l G2 



Also, the value of C sg is chosen so X csg is small at the operating frequency. 



492 Electronic Engineering 

The class C amplifier differs from the small-signal amplifiers in the 
method employed for control grid or base bias. In small-signal tube 
amplifiers, the bias was obtained from a resistance in the cathode circuit. 
This type of bias is satisfactory if low-power stages are considered. How- 
ever, the power lost in the cathode circuit becomes rather large in high- 
power stages. Fortunately, a different means of bias is possible if the 
control grid draws current ; the flow of electrons from the control grid to 
ground through R g (Fig. 12.10) can provide a negative potential on the 
control grid. Accordingly, the size of the control grid resistor R g is found 
by the relationship 

R. = T 2 (12-23) 

where V a is the required grid bias potential and I a is the d-c grid current 
drawn by the grid. The value of C c (Fig. 12.10) must be large enough to 
store the charge produced by the pulse of grid current, so an essentially 
constant current flows through R g between current pulses. If the old rule 
of thumb that X c be equal to approximately 0.lR g is applied, this condition 
is usually fulfilled. This type of bias is known as grid-leak bias. 

The self-bias system described previously has one bad characteristic. 
If the a-c driving signal should be removed from the control grid, the bias 
on this stage will be reduced to zero. Zero bias will allow large currents to 
flow through the tube. These large currents usually destroy the tube in a 
short time. Consequently, protective circuits are usually incorporated in 
this type of circuit to remove the plate supply voltage if the excitation fails. 
Sometimes enough cathode bias or fixed bias is used in conjunction with 
grid-leak bias to protect the tube. Frequently an RF choke is connected in 
series with the grid-leak resistor so essentially no signal currents flow 
through the grid resistor. Otherwise signal power is wasted. 

In the transistor amplifier, the current flowing through the base circuit 
resistance produces reverse bias in the same manner as grid-leak bias in the 
tube amplifier. Fortunately, the direction of base current is proper to 
produce the required polarity of bias voltage. Thus 



Ye 

Ib 



(12.24) 



where R b is the magnitude of bias resistance required (Fig. 12.13). The 
voltage V B is the quiescent base voltage and I B is the average value of base 
current. However, in the transistor, zero bias is also cutoff bias. Therefore, 
no protective devices are necessary, since the collector current reduces to 
essentially zero when the excitation is removed. 

The value of base bias required for class C operation of a transistor 
amplifier cannot be specified in terms of the cutoff bias (as for a vacuum 



Large-Signal Tuned Amplifiers 493 

tube) because the "cutoff" bias voltage of a transistor is actually a forward 
bias. However, the transistor class C bias can be determined in terms of 
the collector current conduction angle (Fig. 12.15). In this figure, an 
operating line is shown on a modified set of constant current characteristics 
(similar to those of Fig. 12.14). The desired bias voltage is V B . The base- 
emitter voltage at which collector current begins to flow is V x and the peak 




Fig. 12.15. The relationships used to determine the transistor base bias voltage as a 
function of conduction angle. 

source voltage, which occurs at the collector saturation point P, is V P . 
From Fig. 12.15 it can be seen that 



V B 

= cos - = 



V X -V R 



V P -V B 2 (V P - V x ) + (V x - V B ) 



(12.25) 



where is the collector current conduction angle. Then, solving for. 

(V P - V x ) cos 0/2 



v x -v B 



or 



V r - V K = 



V x ~V B = 



1 - cos 0/2 



V„- V, 



(12.26) 



(12.27) 



sec 0/2-1 

The potential difference V P - V x may be determined from the input 
characteristics as well as from the modified constant current curves since 
V P = V BE max + i B max /? s . After determining V P , the conduction angle 
is then chosen and the potential difference between V x and V B is readily 
determined by the use of Eq. 12.27. For example, if the typical class C 
conduction angle 120° is chosen, cos 60° = 0.5, sec 60° = 2 and V x — 



V B 



Vp- 



' x- 



494 Electronic Engineering 

In designing transistor Class C amplifiers, the emitter-to-base reverse 
breakdown voltage V EBO must be considered. If the combination of bias 
potential V B and peak base signal voltage V h exceeds this breakdown 
voltage, steps must be taken to protect the transistor. While several 
configurations are possible, the simplest way to protect the transistor 
is to insert a diode in series with the base. The polarity of this diode 
should permit current to flow when forward bias is applied to the diode- 
base combination. Then, when reverse bias is applied, the diode will 
essentially prohibit the flow of reverse current in the base circuit. Of 
course, the reverse voltage breakdown of the diode should be greater than 
the maximum reverse bias applied to the diode-base combination. This 
method of transistor base protection is also used in some of the pulse 
circuits (for example, Fig. 16.53) in Chapter 16. 

PROB. 12.7. The transistor in Fig. 12.13 is the one whose characteristics are 
shown in Fig. 12.11. The V cc supply is 50 v and it is desired to swing the 
collector current to 8a with i B of 300 ma. A conduction angle of 1 10° is desired. 
The resonant frequency is one MHz and the desired circuit Q is 15. Find the 
values of all pertinent voltages and currents (both a-c and d-c), the input power, 
the output power, the efficiency of the circuit and all circuit components. 
Neglect the loss resistance in the coils and assume k = 0.7. The signal source 
resistance is 10 O. Answer: P = 77.5 w, P z = 90.5 w, P b = 0.42 w, diode 
protection of the base is required. 

12.6 THE DESCRIBING FUNCTION 

The technique of pole-zero plots and consequently of root-locus plots 
assumes that all the circuit elements are linear. However, as indicated in 
Section 10.2, large-signal amplifiers are usually not linear. When a 
nonlinear element was encountered, a graphical approach was used. We 
will now consider a technique which can be used to adapt the pole-zero 
method to circuits which contain nonlinear elements. 

As noted previously, the nonlinear elements produce harmonics which 
are not present in the original waveform. Hence, if a fundamental 
sinusoidal waveform is applied to a nonlinear circuit, the output current 
waveform contains the fundamental signal plus various harmonics. The 
describing function approach is based on the premise that the harmonics 
may be neglected without introducing appreciable error. Since tuned 
amplifiers contain resonant circuits, the harmonics are effectively filtered 
out of the response. Consequently, the use of the describing function 
with tuned amplifiers is valid. 

In using the describing function, the general transfer function of a 
network is divided into two components. One component represents the 
transfer function of the linear elements (a conventional function in s, or 



Large-Signal Tuned Amplifiers 



495 



ja> for steady-state alternating current). The second component is known 
as the describing function and represents the nonlinear characteristic. (If 
more than one nonlinearity is present, all nonlinearities are grouped 
together and are considered as a single nonlinear component.) The 
describing function simply relates the fundamental component in the 
output of the nonlinear device to the fundamental component of the 
input to the nonlinear device. Mathematically this concept (Fig. 12.16) is 
expressed as 

G = G& (12.28) 

where G is the total transfer function. G t is the linear transfer function 
and is a function of radian frequency only. ^ is the describing function 





All 

linear 

elements 

Gi 




All 
nonlinear 
elements 

















Fig. 12.16. A graphical representation of Eq. 12.28. 

of the nonlinear element. As we have already found for the class C 
amplifiers, the ratio of the fundamental component of plate current 
to the fundamental component of grid voltage varies as the conduction 
angle (or signal amplitude) varies. Thus ^ is a function of signal amplitude. 
Fortunately, in most class AB, B, or C amplifiers, the relative phase 
between the fundamental component of output current and the funda- 
mental component of input voltage does not change with frequency if the 
tuned circuit is always tuned to the fundamental. However, in some 
nonlinear elements the phase can change with frequency. Thus, although 
the describing functions with which we will be concerned are functions 
of signal amplitude only, the general describing function may be a function 
of amplitude and also of frequency. An example will be used to help 
clarify the describing function concept. 

Example 12.3. Find the describing function for an ideal class C amplifier. The 
transfer function for an ideal amplifier is shown in Fig. 12.17. 

A plot of the output current as a function of time is given in Fig. 12.18. The 
time when the output reaches its peak value will be used as the reference time 
for this analysis. The peak value of the fundamental component of current in the 
output signal can be found from a Fourier analysis of the output wave. Thus 






I COS (Ot d((Dt) 



(12.29) 



496 



Electronic Engineering 




Output 
(collector 
current) 



Input 
(grid volts) 



Fig. 12.17. Ideal Class C amplifier transfer characteristics. 

By reason of the symmetry involved in Fig. 12.18, this equation can also be 

written as 

2 C" 
I pl =-\ icos(otd(wt) (12.30) 

w Jo 

However, the current is zero from 0/2 to w (Fig. 12.18), so Eq. 12.30 becomes 



1 t>\ 



2 r e ' 2 



cos oot d(u>t) 



(12.31) 



The signal current is given by the expression i = /cos cot. But since the axis 
of the current in Fig. 12.18 is displaced from i = by the amount (/cos 0/2), 
the total- expression for i is 

*= / cos tot - I aos 0/2 (12.32) 



M . 




v ■■ 






■\ - 






Otl 














/ 


'-\ 




/! 


\ 


/ 


\ 


C 


/ 


\ 


/ 


\ 


£ 


/ 


\ 


/ 


\ 




f l 


\ 


/ 


\ 




1 i 


\ 


/ 


\ 


3 


/ 


\ 


/ 


\ 


o 


! 


\ 




/ 


\ 




-A 


e 

2 


*— 


Time 


>- 




-- fi 












<■ V 







Fig. 12.18. The output current of a class C amplifier. 



Large-Signal Tuned Amplifiers 

/ is inse 

= - I (/cos cof - /cos 8/2) cos mt d{(ot) 

T JO 

ted to ) 

re i 1 

---sinej 



When this value of / is inserted into Eq. 12.31, I Pl is 

["' 

This integral can be evaluated to yield 

2/ 



/j>i — 



or 



Zff 



497 



(12.33) 



(12.34) 



(12.35) 



Note that if the operating point in Fig. 12.18 is moved until the device is 
operating class A, the ratio of / (the peak output current) to the peak input 



l.U 








































0.8 




















0.6 






































0.4 


























































0.2 




















n 





















40° 80' 120° 160" 200° 240* 280" 320* 360* 
Conduction angle in degrees 

Fig. 12.19. A plot of 'SjK for the ideal Class C amplifier. 

voltage V t is the midband gain of the amplifier K. Thus, if both sides of Eq. 12.35 
are divided by V u the equation becomes 



*J± = &=^(8 -sine) 
V t 2w 



(12.36) 



A plot of "S\K for various values of 6 is given in Fig. 12.19. 

From this example, we note that a class C amplifier can be treated 



much as a linear amplifier if 



G = G, 



(12.28) 



where G is the total transfer function of a stage. The describing function 



498 Electronic Engineering 

^ is obtained from Eq. 12.36 and is used to replace the reference gain K 
of the amplifier. The term G t contains the transfer function for the 
coupling circuits, including the tuned circuits. Thus, the treatment of 
large-signal amplifiers is similar to that of small-signal amplifiers except K 
is replaced by the describing function term &. 

One word of caution should be given at this time. The ideal class C 
amplifier output current given in Fig. 12.18, and the actual class C 
amplifier output current (Fig. 12.8) may not agree. When flattening or a 
dip occurs in the output current (due to a high amount of input drive), 
the actual value of ^ will be less than that shown in Fig. 12.19. However, 
if the waveform is known, (or can be determined), the correct value of ^ 
can be calculated. 

The foregoing example developed the total transfer function IJV g for 
a vacuum tube. When the current gain of a transistor is considered, an 
interesting situation occurs. Since both base current and collector 
current begin to flow at approximately the same base-to-emitter voltage, 
the conduction angle of base current and collector current is essentially the 
same. Thus, the ^ of current gain is s=» 1 unless saturation occurs in the 
collector circuit. However, the ^ for the ratio of base current to base 
voltage will vary in a manner similar to that shown in Fig. 12.19. 

PROB. 12.8. Find the value of ^ for the amplifier in Example 12.1. 
PROB. 12.9. Find the value of S? for the amplifier in Prob. 12.7. Answer: 
& for IJI b is 0.97, & for /JV S = 0.505. 

PROB. 12.10. A class C amplifier is connected as shown in Fig. 12.13. The 
conduction angle is 120°, h u is 50, and the effective shunt resistance (including 
that coupled back from the 100 Q resistor) for the Z.j and C circuit is 2 Kfl. 
Assume that \\h M » 2 KQ. Find the current gain of this transistor. (Assume 
that the output current is the current through C and L x rather than the current 
in the 100 Q, resistor.) 

12.7 EFFICIENCY OF CLASS C AMPLIFIERS 

The high efficiency of the class C amplifier can be understood by 
reference to Fig. 12.20. The instantaneous power the plate or collector 
must dissipate is equal to the instantaneous value of plate (or collector) 
voltage times the instantaneous value of current. Since current only flows 
while the plate voltage is low, the plate or collector dissipation is relatively 
low. However, the total power into the stage is equal to the V vr supply 
(or collector supply) voltage multiplied by the average current. The 
difference between the input power and the plate or collector dissipation 
is the useful output power. 

The theoretical maximum collector or plate efficiency can be calculated 
in the following manner. From Fig. 12.20, the average power into the 



Large-Signal Tuned Amplifiers 



circuit is 



F < = i \'v cc i(t)d(cot) 

2n J-v 



499 



(12.37) 



where V cc is the collector supply voltage. (This voltage would be V PP for 
a tube circuit), and i(f) is the current as a function of time. Since the 
waveforms are symmetrical about the reference point (Ref. of Fig. 12.20) 
Eq. 12.37 can be written as 

P, = £- f V cc i(0 d(mt) 

ZTT JO 



(12.38) 



Voltage - 



i/S-^v, 

/ 1 \ 


1 „ 


— e— 


1 -w 


K — 





Fig. 12.20. A plot of plate or collector current and voltage in a Class C amplifier. 

Now, if the current i(f) is assumed to be a section of a sinusoid within the 
conduction angle 0, i(t) can be written as 

/(/) = /(cos mt - cos 0/2) (12.39) 

Notice that 7(1 - cos 0/2) is the peak value of current in the collector 
circuit and that i(/) is zero at tot = 6/2. Equation 12.38 can now be written 

as 

P . = Ysd ((cos cot - cos 9/2) d(cot) (12.40) 

7T JO 

or 



P. = 



VccI 



[sin (0/2) - (6/2) cos (0/2)] 



(12.41) 



Although the actual collector (or plate) voltage cannot swing all the way 
to zero in an actual amplifier, this condition would still give the limiting 
value of power output. Accordingly, the average collector dissipation, P d , 
can be written as 

p d = — f * (V cc - V cc cos mt)i(t) d(tot) (12.42) 

2tt J—tt 

The term (V cc - V cc cos a>i) is the relationship which describes the 
voltage on the collector. Now, if the point Ref. of Fig. 12.20 is taken as 



500 Electronic Engineering 

reference and it is observed that i(t) = for it > t > 0/2, Eq. 12.42 
can be written as 

I rw 
P a = ~ V cc (l — cos cof)/(cos cot - cos 0/2) d(cof) (12.43) 

•7T JO 

The power output P of the stage is given by the relationship 



P = P t -P d = 



VccI 



' re/2 p/2 

cos 2 cot d(cot) — cos 0/2 cos cot d(cot) 
.Jo Jo 



(12.44) 
or 

P = — (0 - sin 0) (12.45) 

4-n- 

Therefore, the maximum collector efficiency v\ v of a class C amplifier 
with a conduction angle is 

P, F C c^/^(sin 0/2 - 0/2 cos 0/2) 

— sin 

Vp ~ 4 sin 0/2 - 20 cos 0/2 ( ' ' 

When is equal to it, the stage would be operating as a class 5 amplifier. 
As a check on Eq. 12.47, when 6 = w, the maximum efficiency is 0.785 
or 78.5 %. This is the value of maximum efficiency for a class B amplifier 
as found in Chapter 10. As another point of interest, note that as -»■ 0, 
the sin -* 6 and the sin 0/2 -»- 0/2. When these values of are substituted 
into Eq. 12.47, the result is an indeterminate form (0/0). However, 
when L'Hospital's rule is applied, the limit of Eq. 12.47 as 0->-O is 1. 
Hence, the efficiency approaches 100% as 0->-O. However, as 6-+0, 
the power output also approaches zero for a finite collector or plate 
current. Accordingly, a compromise must be made between efficiency and 
output power. 

In the usual design procedure, the collector or plate current is allowed 
to approach the permissible maximum. Then, a rough approximation 
can be found from Eq. 12.45. 

The value of is made as small as possible to keep the efficiency high 
but still produce the required output power. The value of V cc in Eq. 12.45 
is made as high as practical to keep the conduction angle low and conse- 
quently the efficiency high. However, after all these preliminary calcula- 
tions, solutions of the type indicated in Example 12.1 or Section 12.4 must 
be found to verify the behavior of the circuit. 

In many design problems, the stage under development is required to 
produce a certain amount of power at a required voltage level to drive the 



Large-Signal Tuned Amplifiers 

next stage in the amplifier. In these problems, the type of coupling circuit 
must be considered as well as the operating voltages and currents of the 
tube or transistor. As a final word of caution, Eq. 12.47 gives the maximum 
theoretical efficiency of a circuit. Actual circuits will always have efficien- 
cies lower than the values indicated by Eq. 12.47 for a given 6. (Actual 
amplifiers are not driven to collector volts, and the collector current is 
usually not sinusoidal.) 

PROB. 12.11. The characteristics of a 4-65A tube are given in Fig. 12 21. 
Design an amplifier using this tube to drive the tube of Examples 12.1 and 12.2. 
List all voltage and current requirements for this driver stage. 



V f = 6 volts I 
Grid-No. 2 volts = 250 
Grid-No. 2 milliamperes = I G2 
Grid-No. 1 milliamperes = I G1 
Plate milliamperes = I P 




1500 2000 2500 3000 3500 

Plate volts 
Fig. 12.21. Constant current curves for a 4-65A tube. (Courtesy of Radio Corporation 
of America.) 

PROB 12.12. Design a class C power amplifier to operate at 1 MHz using a 
2N1899 transistor with V cc = 50 volts The driving source resistance is .3 ohn*. 
Use in max = 8a, i B max = 0.3a and conduction angle = 120 Determine 
power C output, collector efficiency and driving power. Is base circuit protection 

PROB d 12.13. A 4-1000A tube is to be used as a class C amplifier Refe^e^ to 
a tube manual indicates the proper power output can be achieved if icon rol grid 
bias supply of -200 v and a plate supply of 5500 v is used The screen is main- 
tained at +500 v. If the grid signal is 325 v peak, find the values of currents, 
voltages, powers, and efficiency. 

PROB. 12.14. A triode class C amplifier is analyzed and found to have the 
following voltages and currents ; 

Control grid supply -1000 v 

Plate supply + 10,000 v 

Peak signal plate voltage 9000 v 

Peak signal plate current 2 amp 

Peak signal grid voltage 2000 v 



502 



Electronic Engineering 



Peak signal grid current 0.1 amp 
Average plate current 1 .2 amp 
Average grid current 0.06 amp 

(a) What is the signal power output? 

(b) What is the plate power input ? 

(c) What is the plate efficiency ? 

(d) What is the signal power required to drive the grid circuit ? 

(e) If self-bias is used, what size of grid resistor is required ? 
(/) What impedance must the load in the plate circuit have ? 

PROB. 12.15. A circuit is connected as shown in Fig. 12.13. The characteristics 
of this transistor are given in Fig. 1 2. 1 1 . The internal resistance Z g of the gener- 
ator is 2000 ft and V cc is -20 v. The maximum collector current should not 
exceed 1 50 ma. A conduction angle of 120° is desirable. The resonant frequency 
is to be 500 KHz and the desired circuit Q is 20. Find the values of all a-c and 
d-c voltages and currents, also the input power, the output power, and the 
efficiency of the circuit. If the resistance of the coil is negligible and k = 0.7, 
find the value of circuit elements. Assume /^ » 500 KHz. 
PROB. 12.16. The characteristics of a 9C21 triode are given in Fig. 12.22. 
Design an amplifier using this tube and a V PP supply of 10,000 v. The control 



1200 




10 15 

Plate kilovolts 



Fig. 12.22. Constant current curves for a 9C21 tube. (Courtesy of Radio Corporate 
of America.) r 



grid is to be biased at -500 v and to have a voltage swing of 960-v peak The 
power output should be approximately 27 kw. List all voltages, currents and 
power as well as efficiency. Sketch the circuit and give circuit element values if 
the resonant frequency is 2 MHz and the circuit Q is 20 



13 



Oscillator Circuits 



In this chapter, an electronic oscillator will be denned as a device that 
generates a sinusoidal voltage or current waveform. As with any electronic 
item, a source of d-c power is required for operation. In general, either 
one terminal pair devices such as klystrons, tunnel diodes, and so forth, or 
two-terminal pair devices such as conventional triodes, pentodes, transis- 
tors, etc., can be used as the active elements. The active elements must 
work in conjunction with passive (R, L, and C) networks. In some micro- 
wave devices, the values of R, L, and C are distributed over the circuit 
rather than being separate lumped elements. However, even in these 
instances equivalent circuits of R, L, and C elements can be developed to 
help visualize the action of the device. 

13.1 OSCILLATORS WITH TWO-TERMINAL ACTIVE ELEMENTS 

If the active element in an oscillator contains only two terminals, this 
active element must be connected either in series or in parallel with the 
passive elements of the circuit. Accordingly, the circuit can be represented 
as shown in Fig. 13.1. If the active element behaves as a negative conduct- 
ance, oscillations can occur if this negative conductance of the active 
elements is greater than or equal to the positive conductance of the passive 

503 



504 



Electronic Engineering 



circuit. To determine how oscillations can occur, and at which frequency 
and at what amplitude a given circuit will oscillate, the following procedure 
is followed. First, a polar plot is made of the negative of the admittance of 
the passive circuit as a function of frequency. Second a polar plot is made 
of the admittance of the active element as some circuit parameter is varied. 

















Active 
element 




Passive 

circuit 

(R, L, and C) 




















Power 
supply 



















(a) 



















Active 
element 




Passive 

element 

<R,L, and C) 




Power 
supply 



















(6) 
Fig. 13.1. A simplified block diagram of an oscillator with a two-terminal active 
element, (a) Series connections; (6) parallel connection. 



This second polar plot is made on the same chart (and to the same scale) as 
the first polar plot. Usually, the circuit parameter which varies the admit- 
tance of the active element is the magnitude of the a-c signal in the circuit, 
although other circuit parameters can be used. Wherever the two plots 
intersect, the conditions are suitable for the circuit to oscillate. The 
frequency of oscillation and usually the magnitude of oscillation can be 
determined from the plot. 

The reader was introduced to the tunnel diode in Chapter 3 (Section 3.8). 
To illustrate the method just discussed, an example will be given for a 
tunnel diode oscillator. 



Oscillator Circuits 



505 



1 

b 



1.0 ma 




— h 


.— Slope = —g 




Iv 


T 


1 


1 ll 1 


1 


1 


— h 

Vp 




1 


0.5 



Diode voltage in volts 
Fig. 13.2. The characteristic curve of a tunnel diode. 

Example 13.1. The characteristic curve of a tunnel diode is given in Fig. 13.2. 
This tunnel diode is connected as shown in Fig. 13.3a. The diode specifications 
list a total shunt capacitance C d of 7 pf. (An inductance of 6 x 10" 9 h is also 
present but will be ignored in this example.) Find the frequency of oscillation 
and the magnitude of voltage across the tuned circuit for this configuration. 1 If 
the resistance of L is ignored, the tunnel diode will be biased at +0.12 v. Thus 
the tunnel diode can be replaced by an equivalent circuit (consisting of a negative 
conductance —g and capacitance C a ) as shown in Fig. 13.36. 



Equivalent circuit 
for the tunnel diode ^ 




(a) (b) 

Fig. 13.3. A tunnel diode oscillator, (a) Actual circuit; (b) equivalent circuit. 

1 This circuit is idealized to the extent that a battery of 0.12 v is not practical. How- 
ever, this voltage can be obtained from power supplies or resistor-battery-capacitor 
combinations. Unfortunately, great care should be taken or the diode will oscillate 
(or switch) with the battery circuit instead of with the tuned circuit. Additional tunnel 
diode oscillator circuits are given in "Designing Tunnel Diode Oscillators," by Wen- 
Hsiung Ko in Electronics, February 10, 1961 ; Vol. 34, No. 6, pp. 68-72. 



506 



Electronic Engineering 




The passive portion of the circuit (includ- 
ing the capacitance Q) will have the 
configuration shown in Fig. 13.4. The ad- 
mittance of the circuit of Fig. 13.4 can be 
written as 

Y(s) = sC + ^ + 1 (13.1) 



Fig. 13.4. The equivalent a-c circuit 
for the passive portion of the circuit 
in Fig. 13.3. 



or 



Y(s) = c **+w«y+wo (13-2) 



When the values of R, L, and C from Fig. 13.4 are substituted into Eq. 13.2, 

v 5 2 + j x 7.4 x 10 7 + 10 12 



Y(s) = 10- 



(13.3) 



Now, for steady-state alternating current, the complex frequency s becomes 
jca in Eq. 13.3. Accordingly 

rO» = 10-xo -" 2 + * 7A * 10 > + 10 " (13.4) 

Therefore, a polar plot of Y(jco) as <o varies is as given in Fig. 13.5. 

The next step requires the construction of a curve which represents the 
admittance rf the tunnel diode. The characteristic curve of the tunnel diode has 
been reproduced in Fig. 13.6. Since the diode is biased at +0.12 v, the Q point 
would be located as shown in Fig. 13.6. The slope of the characteristic curve at 
Q is indicated by the line AB. The slope of this line AB is the admittance of the 
diode with no a-c signal 

0.6 x 10- 3 a 
-*= 0.08v =7.5x10-3 mhos 

This value of g is plotted as point A in Fig. 13.7. 



-g 



+JB 



Y(joi)- 



-7.4 x 10"^ 



-jB 



(a increasing 



-co = 10° 



+g 



co decreasing 



Fig. 13.5. A plot of Y(jw) from Eq. 13.4. 



Oscillator Circuits 



507 



1.0 ma — 




Fig. 13.6. The effect of signal level on the admittance of a tunnel diode. 

If an a-c signal is superimposed on the voltage at Q, the voltage varies along 
the line AB (Fig. 13.6) for small signals. However, as the a-c signal increases in 
magnitude, the voltage must follow the characteristic curve. Accordingly, as 
the signal becomes larger, the voltage digresses from the AB curve. As the a-c 
voltage becomes larger, the average slope* of dljdV decreases to the value 
indicated by the line CD. The slope of this line indicates that the average 
admittance of the diode with an a-c signal of 0.05 v is about 



S 



1.2 x 10- 3 a 
0.25v 



4.8 x lQ-^mhos 



This value of g is plotted on Fig. 13.7 as point B. As the magnitude of the a-c 



7.5 x 10 ~ 3 mhos 

H — ^ H 



-8 



Zero a-c 






4.8 x 10 ~ 3 mhos 



IT 



+JB 



Increases withy 
increasing a-c 






-jB 
Fig. 13.7. A plot of g vs signal amplitude for the tunnel diode of Fig. 13.3. 

2 The idea of using the average slope may cause the reader to question the validity of 
this approach. However, this method is an extension of the describing function noted in 
Section 12.6. 



508 



Electronic Engineering 



signal increases, the value of — g decreases. In fact, for an a-c signal of about 
0.09 v, the average g is about zero. If the a-c signal increases beyond this value, 
the average g actually becomes positive. Accordingly, the plot of g for the 
tunnel diode is as shown in Fig. 13.7. 

The plot of Y(ja>) for the passive circuit was given in Fig. 13.5. A plot of 
— Y(jco) and the plot of g vs signal amplitude (Fig. 13.7) are reproduced on a 
single set of coordinate axes in Fig. 13.8. The intersection of the two plots, 
point P, is the required solution. In this example, the resonant frequency is 10 6 



co increasing 



co = 10 6 radians/sec 
~8 r 



-Y(ju) 



B 

-7.4 x 10~ 3 - 

- 7.5 x 10" 3 - 
(mho) 



+JB 



jL 



Plot of g for the 
tunnel diode 



-+g 



Increasing a-c signal 



-JB 



Fig. 13.8. A plot of - Y(jcu) from Fig. 13.5 and the plot f(g) from Fig. 13.7. 



radians/sec and the magnitude of signal for — g = 7.4 x 10~ 3 mhos is about 
0.04 v. This is the value of a-c voltage for which the operating line just begins 
to leave the line A-B in Fig. 13.6. Hence g is just below 7.5 x 10~ 3 mhos. 

In the foregoing example, if the circuit were at quiescent conditions 
and the a-c signal were zero, the tunnel diode conductance would have the 
value given by point A of Fig. 13.8. Under these conditions, the negative 
conductance of the tunnel diode would be greater than the positive con- 
ductance of this resonant circuit. Therefore, oscillations would build up 
in the circuit until the a-c signal was large enough to shift the value of 
negative conductance in the tunnel diode to point P of Fig. 13.8. At point 
P, the net conductance is zero. Hence, a pole-zero plot would show the 
poles on the yew axis. Therefore, the damping is zero and steady-state 
oscillations occur. Under these conditions, the circuit is in equilibrium and 
will continue operation at this signal level. 

The admittance plots of many devices are more complicated than the 
straight-line plot of Fig. 13.7. In addition, the plots of Y(jw) for many 



Oscillator Circuits 



509 



passive circuits are more involved than the straight-line plot of Fig. 13.5. 
Even so, the method outlined in Example 13.1 can be applied to these 
more complicated circuits. 

PROB. 13.1. Determine the slope of the characteristic curve of Fig. 13.2 at 
diode voltages of 0.05, 0.06, 0.07 . . . 0.18, 0.19 v. Find the average of these 
slopes to determine the value of -g when the a-c signal of Example 13.1 is 
0.05 v rms. 

13.2 OSCILLATORS WITH FOUR-TERMINAL ACTIVE ELEMENTS 

Many oscillator circuits use triodes, pentodes, transistors, and so on as 
the active elements. In all these oscillators, energy of the proper magnitude 
and phase is fed from the output back to the input circuit. This type of 



*i 



"^ * 2 -. 


G23 




* 3 > 


J * 






, 










G34 













*4 



Fig. 13.9. A typical feedback circuit. 

oscillator is a type of feedback amplifier. To visualize the requirements of 
an oscillator, consider the typical feedback circuit shown in Fig. 13.9. 
The transfer function G 13 was found in Chapter 11 to be 

Go* 



r _ 12 _ 

Ul! = 



(11.6) 



x x 1 + G23G34 

In this equation, the feedback was negative feedback. If the feedback is 
positive, Eq. 1 1 .6 becomes 

(13.5) 



Gia = -r 



X x 1 — G23G34 

Whereas most oscillators use positive feedback, some oscillators actually 
require negative feedback for proper operation. In fact, as already noted 
in Chapter 11, an ordinary amplifier with negative feedback may become 
unstable and oscillate. 

If the circuit is an oscillator, the signal x 1 must be zero and the signal x z 
rnust be finite. Accordingly, from Eq. 13.5, 

Gog 



or 



— = 00 
1 — G23G34 

1 — G 23 G 3 4 = 



(13.6) 
(13.7) 



510 Electronic Engineering 

Hence, 

G 23 G 3 4 = 1 (13.8) 

This criterion just establishes the instability of the circuit. In addition to 
this criterion, in order to fulfill our definition of an oscillator the output 
should be sinusoidal. 

The root-locus plot, which was discussed in Chapter 1 1 , is a solution of 
Eq. 13.7 as the gain K of the circuit is adjusted. (The rules for a root-locus 
plot with positive feedback are given in Appendix II.) Accordingly, the 
point of operation must lie on the root-locus plot. In addition, if the 
output of the circuit is to be a steady-state sinusoidal waveform, the point 
of operation must lie on they'w axis of the j-plane. Consequently, the root- 
locus plot of a circuit must cross the ju> axis in the s-plane if the circuit 
is to be used as an oscillator. In addition to this restriction, no other poles 
of G 13 can be in the right half of the s-plane for the value of K which pro- 
duces oscillation. This restriction must be enforced to prevent the 
instability of a pole in the right half plane from "swamping" out the 
oscillation. 

PROB. 13.2. Prove that if x 2 = x x + x t (positive feedback) in Fig. 13.9, then 

G — G 23 

1 — G 23 G 34 



13.3 TYPICAL R-C OSCILLATORS 

Several types of R-C oscillators are in common use. The most common 
type of R-C oscillator (a circuit that is used in many commercial audio 
oscillators) is the type shown in Fig. 13.10. In effect, this circuit is a two- 
stage amplifier with the fe edback loop composed of R u C x , R 2 , and C 2 . 
Triode tubes have been shown, but pentode tubes or transistors could be 
used. With transistor circuits, the parallel combination of R 2 and R in 
would be used in place of R 2 in the following derivation. The value of C c is 
chosen large enough to pass the lowest frequencies to be generated with 
negligible phase shift. 

The voltage feedback in this particular circuit is positive, since there are 
two polarity reversals in the amplifier. Accordingly, the circuit in Fig. 1 3 . 1 
can be redrawn as shown in Fig. 13.1 1. In this figure, G 23 is the gain of the 
two-stage amplifier. Hence, let G 23 be given as K. The transfer function 
G 34 , can be written as 

G Xi R 2 Q/sC 2 )l[R 2 f (l/sC 2 )] 

34 x 3 R t + 1/ S Q + R 2 (l/sC 2 )l[R 2 + (l/sC 2 )] 



Oscillator Circuits 



511 




Fig. 13.10. An R-C bridge oscillator. 



This equation can be simplified to 
1 



G*A = 



5-I-5 — ■ — -p 



(13.10) 



Equation 13.10 can be further simplified by letting R 1 = R 2 = R and 
d = C 2 = C. Then, 

(13.11) 



G34 = 



1 



RC s* + s± + 



1 



KC R 2 C* 

Since this circuit has positive feedback, Eq. 13.5 applies. 

G«a 



G, 3 = 



1 — G 23 G S 



(13.5) 



* — A 



P 



S 2 



Amplifier 
gain = K 



*3 



14 



r 



HVWH 



c 2 



^(-WvW- 



Fig. 13.11. An equivalent circuit for Fig. 13.10. 



512 



Electronic Engineering 



When K (the amplifier gain) is substituted for G 23 and Eq. 13.11 is 
substituted for G 3i , Eq. 13.5 becomes 

G 13 = ^— ; (13.12) 



RC s 2 + s(3jRC) + 1/R 2 C 2 

The root-locus plot for Eq. 13.11 is given in Fig. 13.12. From Fig. 13.12, 
it is obvious that the circuit in Fig. 13.10 can be used as an oscillator. 

The frequency of oscillation as well as the required amplifier gain for 
oscillation can be found either graphically or analytically. In the graphical 







3--V5 _ 








2RC ^X^ 










^* 


RC \. 

+ ) 
















3+^/5 










2RC 







Fig. 13.12. The root locus for the circuit of Fig. 13.10. 

analysis, careful construction of the root-locus plot is required. A spirule 
simplifies finding the required value of K. In some cases, the problem is so 
complicated that the analytical approach is rather difficult to evaluate. 
Hence, the graphical approach is the logical method of solution. However, 
the graphical approach can be very time consuming. 

In this particular problem, the analytical approach is rather simple. The 
first step is to multiply both numerator and denominator of Eq. 13.12 
by (s 2 + s(3/RC) + \jR 2 C 2 ). Then, Eq. 13.12 becomes 

s 2 + s(3/RC) + \jR 2 C 2 



G13 = K ■ 



s 2 + s[(3 - K)jRC] + \jR 2 C 2 



This equation can be written as 



K 



s 2 + 2£ 1 a> B s + oi 2 



(13.13) 



(13.14) 



s 2 + 2i 2 m n s + w 2 

Referring back to Fig. 13.12, we see that the two roots of the denominator 
must occur at +j(» 1 and —j(o x (if the circuit is an oscillator). Accordingly, 
the denominator of G 13 must have the value 

(s + j'oj^s — y'ft>i) = denominator of G 13 (13.15) 



Oscillator Circuits 513 

or 

j* + m * = denominator of G 13 (13.16) 

When Eq. 13.16 is equated to the denominator of Eq. 13.14, the following 

relationship results. 

S 2 + Wl t = s * + 2£ 2 (o n s + co n 2 (13.17) 

Thus, if there are to be steady-state oscillations, the damping factor £ 2 must 
have a value of zero. When the values of £ 2 and co n from Eq. 13.13 are 
substituted into Eq. 13.17, the coefficients of like powers of s can be 
equated to yield 

= ?— ^ (13-18) 

RC 



FromEq. 13.18, 
and from Eq. 13.19, 



a, 2 L_ (13.19) 

Wl _ K 2 C 2 

K =3 (13-20) 



co, = — (13-21) 

RC 



Figure 13.12 illustrates that the gain of the amplifier in an oscillator 
circuit must be maintained at a constant value. Figure 13.12 shows that a 
decrease of gain will shift the poles from the jeo axis into the left half plane. 
Under these conditions, the signal output will be an exponentially decaying 
sinusoidal wave. In contrast, an increase of gain will shift the poles from 
thejco axis into the right half plane. In this case, the output waveform is an 
exponentially increasing sinusoidal wave. Consequently, it is important 
for the gain of the amplifier to be maintained at a constant level. 

Many circuits have a sort of "built-in" gain stabilization factor. We 
can visualize this type of gain stabilization by referring to the typical 
dynamic transfer characteristic in Fig. 13.13. For small a-c signals, the 
operation is along the steepest part of the dynamic curve indicated by the 
line AB. However, as the a-c signal increases in magnitude, operation 
extends into the less steep portions of the dynamic transfer characteristic 
and the average slope decreases, as indicated by the line CD. Consequently, 
the average gain AK p /AK e decreases as the a-c signal level increases. 

The typical linear oscillator circuits are adjusted so the gain is slightly 
higher than the gain necessary for oscillations with very small signals. 
Hence, the poles in the root-locus plot are slightly to the right of thejco 
axis (in the right half plane) when the oscillator is first turned on. Conse- 
quently, the level of signal increases. As the signal level increases, the gain 



"4 Electronic Engineering 

of the amplifier decreases and the poles of the root-locus plot slide back 
to thejw axis. If the signal increases or decreases beyond this value, the 
gain of the circuit will change in such a direction as to return the operating 
point back to the ju> axis. Unfortunately, this method of gain control 
causes distortion in the output signal because of operation into the 
non-linear portion of the plate characteristic curves. 

The circuit of Fig. 13.10 has a device added to the circuit to maintain the 
gain of the circuit at a nearly constant level. This device is the lamp in the 
cathode circuit of the tube V x . This lamp (with tungsten filament) acts as an 




C A 



Grid voltage 
Fig. 13.13. A typical dynamic plate characteristic curve. 



unbypassed cathode resistor. Hence, the lamp produces negative feedback 
in the circuit. For low output, the current through the lamp is low. Be- 
cause of the temperature dependence of the lamp filament resistance, the 
low current through the filament allows the filament to present a rather low 
resistance and consequently a low value of negative feedback. In contrast, 
for high output, the current through the lamp filament is high. This high 
current raises the temperature and consequently the resistance of the lamp 
filament. Hence, increased output causes increased negative feedback in 
the circuit to help stabilize the gain. Thus, the operation can be in the 
linear portion of the tube characteristics and still have gain control. 
Consequently, the signal output has very little distortion present. 

The circuit of Fig. 13.14 is an improved version of the oscillator of 
Fig. 13.10. In Fig. 13.14, the forward gain of the amplifier is completely 
controlled by negative feedback. The ne gative feedback path consists of 
the resistor R f and the lamp which acts to stabilize the gain . In addition to 
increased gain stability, the additional negative feedback improves the 



Oscillator Circuits 



515 




Output 



Fig. 13.14. An R-C oscillator with negative feedback gain control. 

waveform of the output signal. The negative feedback circuit must be 
designed so that the gain K f = 3 as given by Eq. 13.20. 

The R-C oscillator configuration of Fig. 13.14 is commonly known as a 
Wein-Bridge oscillator because of its bridge characteristics, which are 
illustrated in Fig. 13.15. The biasing resistor R 3 has not been included in 
this figure because its resistance is (and needs to be) high in comparison 
with the impedance of the R& branch. Also, C F has been omitted 
because its only purpose is to block direct current. Observe that the 
amplifier input is between nodes A and B and the amplifier output is 




Fig. 13.15. Illustration of the bridge circuit arrangement of the oscillator of Fig. 13.15. 



516 



Electronic Engineering 



applied across the opposite nodes. Since the voltage gain of the two- 
stage amplifier is of the order of thousands, the bridge is very nearly 
balanced in normal operation. Also note that the transistor has little 
loading effect on either the positive or negative feedback circuits because 
of the small current which flows in the branch AB in comparison with the 
currents in the other branches. 

PROB. 13.3. The circuit of Fig. 13.14 is to generate sinusoidal signals between 
100 and 1000 Hz. The collector resistor R C2 is 2.2 KO. The lamp is rated 3 w 
at 115 v and has about 2 KO resistance in this application. The transistors are 
type 2N 2712. Determine suitable values for all circuit components which have 
not been given, including the range of capacitance C\ = C 2 . Ri should be several 
times as large as R C2 but small compared to R 3 . V cc = 20 v. 



°V„ 




Fig. 13.16. A three-stage R-C oscillator. 

PROB. 13.4. An R-C oscillator is shown in Fig. 13.16. (a) Find the required 
transfer function, (b) Draw the root-locus plot, (c) Find the required circuit 
gain and the resonant frequency. Neglect the stray shunt capitance. Assume 
each tube and load resistor forms an amplifier. 



13.4 L-C OSCILLATORS 

R-C oscillators are used almost exclusively in the frequency range 
below about 500 KHz. However, the shunt capacitance in the R-C coupled 
stages becomes troublesome at higher frequencies and tuned amplifiers 
which utilize this shunt capacitance as part of the tuning capacitance, as 
discussed in Chapter 8, are used as the basic amplifying device. In fact, 
an amplifier which has capacitive coupling between the input and output 
circuits may oscillate if both the input and load circuits are inductive. 
The conditions required for oscillation were discussed in Section 8.6. 
The oscillator which results when these conditions are intentionally met 



Oscillator Circuits 



517 



oVc, 




Equivalent 

resistance of 

the actual 

load 



Fig. 13.17. A tuned-collector oscillator. 

is known as a tuned-plate tuned-grid (or tuned-collector tuned-base) 
oscillator. 

The single-tuned inductively coupled amplifier can be used as an 
oscillator if some of the energy in the output circuit is coupled back to the 
input circuit, as shown in Fig. 13.17. 

The pole-zero plot for the voltage gain of the tuned-collector amplifier 
was developed in Chapter 8 and is given in Fig. 13.18. The root-locus 
plot for positive feedback is also given in this figure. Observe that the 
frequency of oscillation is very nearly the resonant frequency of the 
tuned circuit if the circuit Q is high because the amplifier (open-loop) 




Fig. 13.18. Pole-zero and root-locus plot for the tuned-collector oscillator. 



518 



Electronic Engineering 



poles are very near they'co axis. The actual frequency of oscillation and 
the mutual inductance requirements may be determined from an analysis 
of the equivalent circuit of Fig. 13.19. Small-signal operation is assumed 
initially because the equivalent circuit is valid only for small signals. 
Since the transformer primary impedance is ja>L x + {wM) 2 jR i (Eq. 8.82), 
assuming R ( » coL 2 , the nodal equation for the collector node may be 
written 



g m v = 



G + jcoC + r 



1 



jcoL, + (a>M) 2 /Rj 
where G includes the conductance of the load resistance R L . 



(13.22) 



g m V 



© 



C=±r 



M 



Li. 



Fig. 13.19. An equivalent circuit for the tuned-collector oscillator. 

For the hybrid-77 equivalent circuit, the voltage V is approximately equal 
to the base-emitter voltage, providing that r„ is small and the frequency 
of operation is not appreciably above f p . Also, if R t » coL 2 , as previously 
assumed, the base-emitter voltage is essentially jwMI p , where /„ is the 
signal current through L x . Then 

V= ja>MV ° . (13.23) 



jwLi + 



(coM) 2 



The voltage gain with positive feedback may now be written using 
G 23 = VJV, which is obtainable from Eq. 13.22, and G 34 = VjV from 
Eq. 13.23. 

G 23 _ gJ[G + jcoC + RtKjwLiRj + co 2 M 2 )] 
Jt"Mg m 



Cl3 = 



1 — GooG, 



(13.24) 



1 - 



C/toLx + w 2 M7/?,)(G + jcoC) + 1 
Oscillation occurs when the denominator of Eq. 13.24 is zero. Then 



• r ^ 2r „ o> 2 M 2 G ja) 3 M 2 C , 

jmLfi - m % L x C + — + J ~— + 1 - ja>Mg m = 

R* Rj 



(13.25) 



Oscillator Circuits 519 

Equating the real terms to zero 



2r „ co 2 M 2 G 



or 



co 2 L x C = ^jzl^. + i (13.26) 



1 +^- 03.27) 



T „ M 2 G L X C L 2 C% 
R t 

Since l/^C = co n 2 , where a>„ is the undamped resonant frequency of the 
tuned circuit, 

» = »Jl + <**&* (13.28) 

Equation 13.28 shows that the frequency of oscillation depends on the 
load conductance G, the input resistance R t , and the coupling impedance 
co n M. The oscillation frequency will be very nearly co n if (a> n M ) 2 GjR t is 
small in comparison with unity, which is true if the circuit Q is high. 

The required value of mutual inductance may be obtained by equating 
the imaginary terms of Eq. 13.25 to zero. Then 

^~ - <oMg m + coLfi = (13.29) 

or 

M 2 - ^f^ + ^ = (13-30) 

to C m C 

Using the quadratic formula, 



M = ^±7(^T-^ (13.31) 

* = f# ± y^f) <>332> 



If the circuit Q is high, co 2 ~ \\L X C. Then 



The solution which results from the positive sign preceding the radical 
in Eq. 13.33 may be discarded because it yields unrealizable values of M. 
Normally, 4Glg m 2 R i « 1 ; therefore, the approximation (1 — x) XA ~ 
1 — x/2 can be used to simplify Eq. 13.33. Then, 

M ~ ^ (13.34) 



520 



Electronic Engineering 



Note that the required value of M is independent of frequency, provided 
that a variable capacitor is used and L x remains fixed. The value of M 
obtained from Eq. 13.34 results in class A operation. Larger values of M 
cause increased base drive and result in class B or class C operation, 
depending on M. 

The Hartley oscillator circuit shown in Fig. 13.20 is similar to the 
tuned-collector circuit except a single, tapped coil is used and the tuning 



R 2 



Si- 




»La 



Output 



Fig. 13.20. A Hartley oscillator circuit. 

capacitor tunes the entire coil. In this arrangement, the coupling between 
the collector and base circuits does not depend on the mutual inductance 
between L x and L 2 because an a-c voltage across Lj is also applied across 
the series combination of C and L 2 . Therefore, signal current will flow 
through L 2 and the voltage across L 2 is the feedback voltage to the base- 
emitter junction. The signal currents through L t and L 2 are essentially 
equal (because the tuned circuit current is large compared with the 
collector current), so that the ratio of collector voltage to base voltage is 
essentially LjL 2 . This ratio is the voltage gain, and the reciprocal (L 2 /L x ) 
is the feedback ratio. Thus Kfi = 1, provided that L 2 \L X is large enough 
to cause oscillation. Note that one end of the tuned circuit is at the same 
signal potential as the collector, and the other end of the tuned circuit is 
the same signal potential as the base, and the coil tap is at the same 
signal potential as the emitter. This signal arrangement always holds for 
the Hartley oscillator. Since the input signal is not referenced to ground, 
any one of the three-coil or electrode terminals may be at signal ground 
potential. The oscillator operation is unaltered by the choice of ground 
point except, of course, that the output terminal must not be at the signal 
ground point. An example will be used to illustrate one method of 
designing a Hartley oscillator. 

Example 13.2. A 2N 1613 transistor is to be used in the Hartley oscillator 
circuit of Fig. 13.20. Let us assume the load resistance R L to be 10 Kfi, the 



Oscillator Circuits 521 

transistor output resistance R to be 40 KQ, and the Q of the coil to be 100. 
The oscillator frequency is to be 1 .0 MHz. The circuit Q should be high to insure 
good frequency stability. In this example, we will design for Q = 50. The base 
driving power is very small in comparison with the power furnished to the load 
or dissipated in the tuned circuit and, therefore, its effect on the circuit Q will 
be neglected. Then the parallel combination of R L and R is R x = 8 KQ. 

The inductance L x can be determined by the following method. The total 
shunt collector circuit resistance is (using Eq. 8.9): 

R sh = Qm^ (13.35) 

The portion of this shunt resistance contributed by the coil resistance in the 
tuned circuit is 

Rpar = Qo<*oL X (8.8) 

But R sh is the parallel combination of R mr and R x . Then, using Eqs. 8.8 and 
13.35 



RparRsh QoQ 



and 



In this example 



Rpar ~ Rsh Qo ~ Q 

RxiQo - Q) 



o^ (13.36) 



L x = , n " (13.37) 



8xl0 3 (50) 
1 6.28 x 10 6 (100)(50) ' M 

The class A voltage gain of the amplifier is approximately g m R sh , providing that 
O, « (hfe + IK and the oscillator frequency is not appreciably above f p . These 
constraints are met in this example. Now, R sh is 8 KCl in parallel with Q. u> L x = 
8 KQ, or 4 KO. The Q-point collector current will be chosen as 2.0 ma. Then 

G, ^g m Rsn = ^,IcR s n = 80 x 10- 3 (4 x 10 3 ) = 320 

The oscillator will operate class A if 

L 2 = LJG V = 12.7/320 = 0.04 fih 

However, a change in parameters or loading might stop the oscillation in this 
class A mode. The oscillator will be much more dependable if the inductance L 2 
is increased by a factor of at least 4 or 5. The oscillator will then have much 
better amplitude stability and greater power output. Then 

L 2 = 5(0.04) ,uh = 0.2 fih 

The tuning capacitance can be found from the relationship 

1 1 



(o 2 L (6.28 x 10 6 ) 2 (12.9 x lO" 6 ) 



= 1960 pf 



522 



Electronic Engineering 



The bias components are chosen to provide about 2.0 ma quiescent collector 
current and the blocking capacitor C should have reactance equal to approxi- 
mately one-tenth of the bias resistance. 

The Colpitts oscillator shown in Fig. 13.21 is almost identical to the 
Hartley except that the tuned circuit capacitance, instead of the inductance, 
is tapped. Also, an r-f choke has been added to permit the application of 



-°v„ 



[RFC 



i?2- 




^=Ci 



i?l« 



-±r :=f=c 2 



-|( n Output 



\Rl 



H( 

Fig. 13.21. A Colpitts oscillator. 

direct current to the collector and to present a very high impedance at 
the oscillation frequency. The design of a Colpitts oscillator may follow 
the pattern given for the Hartley oscillator in Example 13.2 but with 
jwL x and ju>L 2 replaced by \lja>C x and l/jcoC 2 respectively. 

uses a 2N1613 transistor and 
as the Hartley oscillator of 

2N2844 field effect transistor, 
circuit components if /„ = 1 
resistance R L = 10 K£2. The 
Design for class C operation. 

,3 fih, njn ~ 1J6. 



PROB. 13.5. Design a Colpitts oscillator which 
has the same specifications and load resistance 
Example 13.2. 

PROB. 13.6. Design a Hartley oscillator using a 
Draw a circuit diagram and determine suitable 
MHz, V DD = — 15 volts and the external load 
coil Q = 150 and the desired circuit Q = 100. 
Specify the coil tap point njn. Answer: L x = 5 



13.5 CRYSTAL-CONTROLLED OSCILLATORS 

A general class of oscillators which achieve very good frequency sta- 
bility because of the exploitation of a high Q circuit is the crystal-controlled 
oscillator. In the "crystal" oscillator, the conventional L-C circuit is 
replaced by a quartz crystal. The crystal has the property of producing a 



Oscillator Circuits 523 

potential difference between its parallel faces when the crystal is strained or 

deformed. Conversely, when a potential difference is applied across the 

faces of a crystal, it will deform or change shape. This property, which is 

known as the Piezo-electric effect after its discoverer, causes the crystal to 

behave as a very high g-resonant circuit. The crystal will vibrate readily 

at its mechanical resonant frequency, but because of its associated electrical 

properties the crystal behaves as though it 

were an L-C circuit with extremely high Q 

(of the order of thousands). The crystal is 

cut into very thin slices and then carefully 

ground to the desired resonant frequency. 

The orientation of the slice, with reference to _^ 

the crystal axes, determines the properties of c ~r- Ch ' 

the crystal, such as vigor of oscillation and 

variation of frequency with temperature. 

The equivalent electrical circuit of a crystal 
is given in Fig. 13.22. The crystal itself be- 



haves as a series R-L-C circuit. However, F|g U22 The equivalent 
the electrical connections must be made to c j rcu j t of a crystal mounted 
the crystal faces by conducting electrodes or in a holder, 
plates, known as a crystal holder. The crystal 

holder provides a capacitance, shown as C h in Fig. 13.22, which is in 
parallel with the crystal circuit. Thus the crystal behaves as a series 
resonant circuit at its natural resonant frequency, but at a slightly higher 
frequency the net inductive reactance of the crystal resonates with the 
crystal holder capacitance to produce parallel resonance. The parallel 
resonant frequency is only slightly higher than the series resonant frequency 
because the equivalent inductance of the crystal may be of the order of 
henries. This extremely high equivalent inductance accounts for the ex- 
tremely high Q of the crystal and provides a very impressive rate of 
change of reactance with frequency. 

The reactance of a typical crystal in a holder is sketched as a function 
of frequency in Fig. 13.23. Note that the reactance is inductive only 
between the series resonant frequency <o s and the parallel resonant 
frequency u) p . These frequencies differ by a very small percentage (a few 
hundred Hz per MHz); therefore, the effective inductance changes very 
rapidly with frequency in this region. 

The crystal can replace the tuning inductor in a conventional circuit as 
illustrated by the Colpitts-type circuit of Fig. 13.24. The oscillator may be 
designed as a conventional oscillator and the crystal will provide the 
proper inductance for operation very near its natural resonant frequency. 
A change of tuning capacitance changes the impedance of the tuned 



524 



Electronic Engineering 



JX o 




Fig. 13.23. A sketch of reactance as a function of frequency for a crystal in a holder. 




Fig. 13.24. A Colpitts oscillator with a crystal for a tuned circuit. 





(a) (b) 

Fig. 13.25. Some typical crystal oscillator circuits. 



Oscillator Circuits 



525 



circuit but has little influence on the frequency of oscillation because of 
the compensating change of effective inductance. 

In the circuit of Fig. 13.25a, the crystal operates in its series mode. At 
the resonant frequency of the crystal, the oscillator operates as a Hartley 
circuit. The circuit of Fig. 13.256 uses the crystal in its series mode to 
couple two transistors, one of which is operating in the common base 
configuration and the other in the common collector configuration. Many 
other circuits may be devised or found in the literature. 

The chief disadvantages of crystals are as follows: 

1. They are fragile, especially the high-frequency crystals, and consequently can 
be used only in low-power circuits. 

2. The oscillator frequency cannot be adjusted appreciably. However, the 
parallel or holder capacitance has some effect on the frequency in the parallel 
mode. 

PROB. 13.7. A quartz crystal has the following electrical characteristics: 

L = 3.2 henries 

C = 0.05 pf 

R = 4000fl 
C h = 6/>/ 
(a) Determine the value of/ s and/„ for this crystal, (b) What is the Q of this 
crystal? (c) Design a Colpitts oscillator which uses this crystal and a 2N1613 
transistor. The value of R L is 10 Kfi and the transistor output resistance is 
40 KX!. Answer: (a)f s = 398 KHz, f p = 401 KHz. (b) Q = 2000. 
PROB. 13.8. The phase shift oscillator shown in Fig. 13.26 is a rather common 
type of R-C oscillator. The tube and load resistor form an amplifier and the 
three capacitors C in conjunction with the three resistors R form the feedback 
path. Make a root locus plot for this circuit and determine the required 
amplifier gain and oscillation freq. 




Fig. 13.26. A phase shift oscillator. 

PROB. 13.9. A reflex klystron is used for generating microwave frequencies. A 
typical reflex klystron contains a built-in resonant cavity, which can be repre- 
sented as a parallel-tuned circuit, (a) Determine the value of L, C, and R for 



526 



Electronic Engineering 



this tuned circuit if the resonant frequency is 10 10 Hz and the Q of the circuit 
is 1000. The conductance of the tuned circuit at resonance is 20 micromhos. 
(/>) Make a plot of - Y(ja>) for this tuned circuit. 

An electron beam passes through a gap in the resonant cavity. The electrons 
in this beam are stopped by the electric field of a negative "repeller" electrode 
and repelled back through the gap in the resonant cavity. The electron beam 
interacts with the electric field of the gap to produce a conductance^ in parallel 
with the capacitor of the tuned circuit. A plot of the value of g as a function of N 
is shown in Fig. 13.27. The parameter N is the number of cycles the electric field 



+JB 



-g N=2%\N = 




Fig. 13.27. A plot of electron beam conductance in a reflex klystron. 



across the gap has completed between the time a given reference electron first 
passed through the gap to the time when this same electron returns to the gap. 
(c) At what value of N will oscillations first begin? (d) What is the frequency of 
these oscillations ? The radius of the admittance spiral decreases as the a-c signal 
increases, (e) At what value of N (for < N < 2) will the a-c signal be 
maximum? (/) As N is increased above this value for maximum signal, the 
oscillations will cease. At what value of N will oscillations cease ? (g) What 
frequency corresponds to the value of N in part/? 

PROB. 13.10. Convert the amplifier of Prob. 12.8 into an oscillator. This 
oscillator will provide the excitation for the tube of Examples 12.1 and 12.2. 



14 



Amplitude Modulation 
and Detection 



In the preceding chapters, the goal has been the development of basic 
principles whereby the process of signal amplification may be achieved by 
the use of electronic devices. In the amplifier it is usually desirable that 
the output quantity or response be the same form as the input quantity 
or excitation. This chapter will consider some electronic devices in which 
the form of the signal is intentionally changed. Generally, these devices 
are known as modulators and the process is known as modulation, which 
means "to change." 

Modulators are needed in a great number of electronic systems. For 
example, in a radio transmitter an oscillator generates the basic radio 
frequency signal which is commonly known as the carrier. If this carrier 
were merely amplified and broadcast, there would be no intelligence trans- 
mitted and the system would be useless. Thus, somewhere in the trans- 
mitter the carrier must be changed or modulated by the intelligence which 
is to be transmitted. The intelligence can then be recovered at radio 
receivers by a device known as a detector. The carrier may be changed in 

527 



528 



Electronic Engineering 



any of several ways such as in amplitude or in frequency. In this chapter 
only amplitude modulation will be considered. 

14.1 AMPLITUDE MODULATION 

In order to gain some understanding of the fundamental principles of 
amplitude modulation, a simple special case will be considered in which a 
carrier with maximum amplitude A c and natural frequency a> e will be 




Fig. 14.1. Modulation (a), the carrier; (b) the modulating signal; (c) an amplitude 
modulated carrier; (d) an overmodulated carrier. 



modulated by a sinusoidal (single-frequency) signal which has a natural 
frequency u> m . A sketch of the modulated carrier voltage as a function of 
time, along with the carrier and the modulating signal, is given in Fig. 14.1. 
As shown in this figure, the maximum amplitude variation from the 
unmodulated value is MA C , where M is known as the modulation index. 
When M has a value of one, the amplitude of the modulated wave varies 
between 2A C and zero. In this case, the carrier is said to be 100% modu- 
lated. When M has the value 0.5, the amplitude of the modulated carrier 



Amplitude Modulation and Detection \! ; We,^^, ^ nMuS o«-.tU*- ac ' 

varies between l.5A e and 0.5/4 c . The carrier is then said to be 50% modu- 
lated, and so forth. Observe that if M exceeds unity, the carrier is com- 
pletely interrupted for a time, the envelope (see Fig. 14.1) of the carrier no 
longer has the same form as the modulating signal, and the carrier is said 
to be overmodulated. Overmodulation naturally causes distortion in the 
system. It may be seen from Fig. 14.1 that the voltage of the modulated 
wave may be expressed by Eq. 14.1 

v = A c [l + M cos o> m t] cos (o c t (14.1) 

This expression could have just as well been in terms of current instead of 
voltage. Also Eq. 14.1 would be more general if arbitrary phase angles 
were included in the expression. However, the results would not be altered 
by the increased generality. Expanding Eq. 14.1, we have 

v = A c cos <a e t + MA C cos co c t cos (o m t (14.2) 

Substituting the trigonometric identity cos a cos b = £[ cos (a + b) + 
cos (a — b)] into Eq. 14.2, we see that 

v = A cos (o c t -\ " cos (co,. + cojt + —— e cos (o> c - ft) m )f 

(14.3) 

It may be seen from Eq. 14.3 that the effect of the modulation is to 
produce two new frequencies, which are called side frequencies . The upper 
side frequency is the sum of the carrier frequency and the modulating 
frequency whereas the lower side frequency is the difference between the 
carrier and modulating frequencies. Therefore, a tuned amplifier which is 
called on to amplify a modulated carrier must have silffin'ffllt hanriwidtll t o 
include the side frequencies. Notice that the modulating frequency is not 
included in the modulated wave. In case the modulating signal were 
derived from a symphony orchestra, each frequency component would 
produce a pair of side frequencies. Consequently, the highest frequency 
components present in the modulating signal would determine the re- 
quired bandwidth of the tuned circuits in the radio transmitting and receiv- 
ing equipment. If these tuned circuits have insufficient bandwidth, the 
highest modulating frequencies will not be reproduced by the receiver. 
Collectively, the upper side frequencies are known as the upper sideband 
and the lower side frequencies are known as the lower sideband. 

Any modulating waveform which is represented as a function of time 
may be resolved into frequency components by Fourier analysis. Thus, the 
bandwidth requirements may be determined whenever the waveform of the 
modulating signal is known as a function of time. 



530 



Elettronic Engineering 



Example 14.1. Consider the rectangular pulse shown in Fig. 14.2a to be the 
modulating signal. The pulse duration is t d and the period is T. Using Fourier 
analysis the frequency components of the modulating signal are found as follows. 
The time reference is taken so that only cosine terms will appear in the Fourier 
series. Then 

An = tL/"* (~H dt = rj. Kcos {-¥*) dt (14 - 4) 



td 



^ K- I ' 



Time 



Mill. 



(a) 



(b) 



Fig. 14.2. (a) A rectangular pulse-modulating signal ; (b) the frequency spectrum of a 
rectangular pulse. 



Since the function Kis zero from t d jl < t < T/2 Eq. 14.4 becomes 



then 



4 n*l* /2nn \ 



T4V 
2irTn 



sin 



& 



tdl* 



2V 



sin 



(?) 



(14.5) 



(14.6) 



Multiplying both the numerator and denominator of Eq. 14.6 by t d and 

rearranging, we have 

2Vt d sin (nntJT) 

(14.7) 



^n — 



mrt d jT 



Thus, the frequency components of the modulating wave are an infinite series as 
shown below. 



2Vt d 



'" [-T (2, \ S ' n (-T) 

— i cos I — t\ + 

{■nt d )lT \t'J + 



(2^)/r cos Y'--; 



(14.8) 



The coefficients of these harmonically related frequency components are of 
the form (sin x)jx. These frequency components, which are spaced at the interval 



Amplitude Modulation and Detection 



531 



/ = 1/r, are shown graphically in Fig. 14.2ft. It was assumed in Fig. 14.2 that 
the period 7" is large in comparison with the pulse duration t d . Of course, all the 
components cannot be shown because they form an infinite series. The frequency 
components of the amplitude modulated carrier are shown in Fig. 14.3 along 
with a sketch of the modulated wave. To include all the side frequencies, the 
bandwidth of the amplifiers which amplify this signal must be infinite. Of course, 
an infinite bandwidth is practically impossible to attain, so a compromise must 
be reached. Frequently the accepted compromise is that all the side frequencies 
up to the first zero amplitude component be included, since the frequency com- 



M 



J-L 



Mi l l 



nlL 



■ II 



(fe ~ Utd) fc <fc + Ut d ) 



t d 

(a) W 

Fig. 14.3. (a) A pulse-modulated wave; (b) the frequency components of a pulse- 
modulated wave. 

ponents beyond this point are of rather small amplitude. This choice would 
result in a modulated signal which departs considerably from the ideal rectangu- 
lar shape, but nevertheless it provides a convenient reference point because the 
first zero amplitude modulating component occurs (from Eq. 14.7) at the lowest 
value of n for which 



2 Vt a sin (nntJT) _ 
A »~ T nntJT 

The lowest value of n at which Eq. 14.9 will hold is 

mrt d 



(14.9) 



(14.10) 



or 



n = 



'a 



(14.11) 



Since the frequency components are separated by the frequency 1/r, the width 
of a single sideband would be (from Eq. 14.11) 



!""(*)"£ 



(14.12) 



532 Electronic Engineering 

The bandwidth requirement would then be 2/t d . Wider bandwidth would 
naturally provide better waveform of the modulated signal. 

The bandwidth requirement for a pulse-modulated wave can also be 
determined from the required rise time of the modulation envelope. The 
analysis of the time response of a tuned amplifier, given in Chapter 8, 
showed that the rise time of the modulation envelope is approximately 
4.4/5, where B is the bandwidth of the tuned amplifier in radians per 
second. Therefore, the bandwidth may be determined from the desired 
rise time. 

PROB. 14.1. A 100 MHz carrier is modulated by a rectangular pulse which has 
a duration of 1 /«ec and a repetition rate of 1000 pulses per second (pps). If 
the side frequencies up to the first zero magnitude component on each side of the 
carrier are to be included, what bandwidth will be required of an amplifier for 
this modulated wave? How many side frequencies would be included in this 
case and what circuit Q would be required of an amplifier which incorporates a 
single tuned circuit? What will be the envelope rise time of this amplifier? 

PROB. 14.2. Repeat Prob. 14.1 but include all side frequencies up to the second 
zero component. 

Returning to the case of a single modulating frequency, we can see 
from Eq. 14.3 that when the modulation index M is unity (100% modu- 
lation) the amplitude of either side frequency is one-half that of the carrier. 
Since power is proportional to the square of the amplitude, each side 
frequency will have one-fourth as much power as the carrier and the total 
side frequency power will be one-half as great as the carrier power. Again, 
this relationship holds only for 100% sinusoidal modulation. 

14.2 MODULATING CIRCUITS 

A large variety of circuits may be used to provide amplitude modulation. 
Only a few typical circuits will be included in this work for the purpose of 
illustrating the basic principles. One common method of accomplishing 
modulation is by using the modulating signal to vary the plate or collector 
voltage of a class B or class C amplifier as shown in Fig. 14.4. Since in 
class B or class C operation, the peak amplitude of the signal voltage in the 
output is very nearly equal to the supply voltage, the amplitude of the 
output voltage follows very closely the variation in voltage supplied from 
the modulator. The modulation is said to be linear when the envelope of 
the modulated wave has the same waveform as the modulating signal. 
Observe that the modulation may be linear even thougn the modulated 
amplifier may be very nonlinear so far as the waveform of each r-f cycle is 
concerned. The tuned coupling circuit essentially eliminates the harmonics 



Amplitude Modulation and Detection 533 

of the carrier frequency as well as the modulating frequency so the output 
is a modulated wave of the form of Fig. 14.1c. The capacitor C provides a 
low impedance path to the carrier currents so these currents do not flow 
through the modulation transformer. On the other hand, the capacitor C 
must not bypass the modulating frequencies. 

The power requirement of the modulator as well as the modulator load 
resistance may be determined from the basic current and voltage relation- 
ships of the modulated amplifier. To obtain 100% modulation, the maxi- 
mum value of the modulating voltage V m must be equal to the power 



r-f amp 




"SB? 

ModulatorT 



°V Pf 




Fig. 14.4. Typical circuits for («) plate and (b) collector modulation. 

supply voltage (V PP or V cc ). Under these conditions, the r-f output of 
the modulated amplifier is equal to zero at the negative peak of the 
modulating signal. Then, using the tube circuit as an example, we have 

= V PP (14.13) 



m(max) 



Also, since the average plate current of the modulated amplifier is reduced 
from I P to zero during this negative half-cycle of the modulating signal, 
the maximum value of the modulating signal current is 

(14.14) 



'm(max) 



= /* 



The modulator power output is 



J mod — 



■J. 



'mlmaxl' m(max) 

2 



VppI P 



(14.15) 



Thus, as shown by Eq. 14.15, the power output from the modulator must be 
equal to one-half the power supplied to the r-f amplifier by the power 



"4 Electronic Engineering 

supply. Therefore, the power from the modulator provides the power for 
the sideband frequencies. It was previously shown that the power in the 
side frequencies is one-half the carrier power for this case. 

The effective load on the modulator may be easily determined since the 
maximum amplitudes of both the modulator voltage and current are 
known. 

R L - = -— (14.16) 

'm(mas) 'p 

The value of supply voltage is known and the value of I P may be deter- 
mined from the graphical analysis of the class B or class C amplifier as 
discussed in Chapter 12. Also, I P is easily measured by a d-c meter in the 
plate circuit of the r-f amplifier. 

The turns ratio of the modulation transformer should be chosen so that 
the load resistance determined by E.q. (14.16) will present the desired load 
resistance for the modulating tube or transistor. 

The modulator may typically be any of the power amplifier circuits 
discussed in Chapter 10. Push-pull circuits are usually used to modulate 
high-power amplifiers. For pentode and tetrode r-f amplifiers, it is helpful 
to modulate the screen grid voltage as well as the plate voltage, since the 
plate potential has a very small effect on the magnitude of plate current 
except at low values of plate voltage. Also, the screen current becomes 
high when the plate voltage is low, unless the screen voltage is reduced 
at the same time. 

PROB. 14.3. A class C amplifier which has 1 Kw output and is 80% efficient is 
to be plate modulated. The plate supply voltage V PP = 2000 v. Determine the 
required power output and load resistance for the modulator. The modulator 
is expected to provide 100% modulation. Answer: P mod = 625 w, R L = 3200Q. 
PROB. 14.4. Draw the circuit diagram for a modulated pentode amplifier in 
which both plate and screen grid are modulated. 

The linearity of plate or collector modulation is usually good. In the 
tube circuit, better linearity is attained when the modulated amplifier has 
grid leak bias rather than fixed bias. The improved linearity results from 
the variation of the bias over the modulation cycle. During the negative 
half of the modulation cycle, the grid draws more current because the 
plate potential is reduced. Thus, the bias is increased and assists in the 
reduction of plate current. Conversely, during the positive portion of 
the modulation cycle, the grid current is decreased because of the in- 
creased plate voltage. Consequently, the bias is decreased, thus enhancing 
the plate current increase. This varying bias tends to offset the tendency 
toward flattening of the peaks of the modulated wave due to saturation 
effects. 



Amplitude Modulation and Detection S3S 

Collector modulation presents a special problem in the transistor 
circuit. The base drive must be large enough to saturate the transistor at 
the peak of the modulation cycle in order to provide linear modulation and 
high efficiency. Consequently, the transistor may be highly over-driven 
during the modulation troughs when the collector current should be 
comparatively small. However, excess charge is stored in the base during 
the time the transistor is in saturation and the collector current cannot 
decrease until this excess charge has been removed. As a result of this 
delay, the large values of collector current are not confined to the period 
during which the collector voltage is low and the collector dissipation is 
increased. In fact, the collector current may increase during the portion 
of the collector voltage cycle when the current would normally decrease. 
This enlarged, out-of-phase current may seriously decrease the efficiency 
of the amplifier. Grid leak type, or R-C bias, will curtail the excess 
stored charge because the base current increases rapidly when the transistor 
is driven into saturation and the resulting increased bias decreases the 
excessive base drive. A discussion of collector-saturation and excess 
stored charge is given in Chapter 16. 

The main disadvantage of plate modulation is the large modulating 
power required when the modulated amplifier is of high power. Of course, 
a lower level stage could be modulated in a high-power transmitter and the 
modulated wave could then be amplified. But class C amplifiers are not 
suitable for amplifying modulated waves because the waveform of the 
modulation is not preserved. Of course, class B amplifiers are suitable, 
but their reduced efficiency is a serious handicap when the power level is 
high. 

Fortunately, grid or base modulation requires much less modulating 
power than does plate modulation. Typical grid and base modulation 
circuits are given in Fig. 14.5. The r-f amplifiers could be operated either 
class B or class C. In these circuits, the modulating voltage varies the grid 
(or base) bias as shown in Fig. 14.6. The dynamic transfer characteristic 
for a typical vacuum tube is given in Fig. 14.6a, and the current transfer 
characteristics of a typical transistor are given in Fig. 14.66. For the 
transistor, the effects of the driving source resistance R g , as discussed in 
Chapter 10, are included. In this figure, the source resistances of the carrier 
and the modulator are assumed to be the same, and in practice these 
resistances should be adjusted to approximately the same value. 

In the r-f amplifiers of Fig. 14.6 it is desirable to vary the output current 
from the maximum design value (point b') at the crest of the modulating 
signal to essentially zero (point c') at the trough of the modulating signal. 
Therefore, the r-f amplifier current peaks should be adjusted to the average 
of these two values (point a) when the carrier is unmodulated. It may be 



Electronic Engineering 

seen from Fig. 14.6 that the maximum amplitude of the modulating voltage 
should be the potential difference between point a and either point b or 
point c on the input voltage axis. This potential difference will be called 
V m . Considering the transistor, we know that the bias current varies 
from the value obtained when the carrier is unmodulated (I Ba ) to zero as 



r-f In 

o— 




Modulator 



vc *BB 



(") (b) 

Fig. 14.5. (a) Grid-modulated and (b) base-modulated amplifiers. 

the modulating voltage varies through its negative half-cycle. Therefore, 
the transducer 1 power required from the modulator is 






(14.17) 



As stated previously, the output resistance of the modulator and its 
associated circuits should be approximately the same as the output resist- 
ance of the carrier source and its associated circuits. These resistances 
should have the value which gives best linearity as determined by the 
technique discussed in Chapter 10. 

In the tube amplifier, the grid draws no current when the carrier is 
unmodulated. Hence, the grid draws current and therefore presents a load 
to the modulator only on the positive peaks of the modulation cycle. This 

1- rhe transducer power is the power furnished by the current generator or voltage 
generator in the equivalent circuit of the source. This is the power delivered to the 
internal resistance of the source in addition to the load. 



Amplitude Modulation and Detection 



537 




a* 



3 
■O 
O 



■a 
a 
« 



a. 



-a 

a 



o 
> 






538 Electronic Engineering 

varying load will cause distortion of the modulation envelope unless one of 
the following conditions is met: 

1. The output impedance of the modulator is small in comparison with the 
minimum load resistance which occurs at the modulation peaks. 

2. A loading resistor, which is small in comparison with the minimum load 
resistance caused by the grid current, is placed across the output of the 
modulator. 

Grid or base modulation has two serious disadvantages when compared 
with plate or collector modulation. First, the power output and efficiency 
of the grid modulated amplifier is comparatively low. Since the unmodu- 
lated plate current peaks can be only about half as large as in the plate 
modulated circuit, the power output and efficiency suffer severely. Second, 
the adjustment of the grid (base) modulated amplifier is more critical and a 
high degree of linearity is more difficult to attain. 

Any of the electrodes of a tube or transistor could be used as the 
modulated element. For example, emitter or cathode modulation is very 
similar to base or grid modulation. The main difference is that the emitter 
(cathode) current is much larger than the base (grid) current so the modu- 
lator power must be much greater. On the other hand, the linearity might 
be better, especially in a tube amplifier, because the cathode current is a 
fairly linear function of the modulating voltage. 



14.3 THE MODULATION PROCESS 

By this time, the inquisitive reader should have posed the question, 
"What basic difference exists between the linear amplifier which only 
amplifies the applied signals and the modulator which produces the sum 
and difference of the applied frequencies in addition to their possible 
amplification?" A clue to the answer might be found in Chapter 10 where 
the production of new frequencies called harmonics is considered. In 
that case, it was the nonlinearity of the amplifier parameters that generated 
the new frequencie s. This nonlinear relationship between the output 
current i Q and the input voltage v t can be expressed by the power series 

'o = ^o + <V< + A 2 v? + A 3 Vt 3 + ■•■ (14.18) 

Now, if the input signal v t consists of two sinusoidal signals, for example, 
a carrier frequency and a modulating frequency, then 

v { = A c cos u)J + A m cos ca m t (14.19) 



Amplitude Modulation and Detection 539 

Substituting this value of v t into Eq. 14.18, we have 

«'o = A o + A Mc cos w c t + A m cos o)J) + A 2 (A C cos ay + A m cos o>Jf 
+ A 3 (A C cos w c t + A m cos wj) 3 + ••• (14.20) 

If we assume that the nonlinearity is such that the fourth and all higher 
order terms are negligibly small, Eq. 14.20 can be expanded to 

'O = A + A \ A c cos <V + A \ A m cos w mt + A % A c cos2 "V 

+ A 2 A m 2 cos 2 w m t + 2A 2 A c A m cos co c t cos m m t 
+ A 3 A C 3 cos 3 a> e t + 3A 3 A c 2 A m cos 2 co c t cos m m t 
+ 3A 3 A c A m 2 cos a> c t cos 2 u> m t + A 3 A m 3 cos 3 u> m t (14.21) 

Most of the terms of Eq. 14.21 have a familiar form. As previously 
discussed, the A term is merely a bias term and the next two terms in- 
volving A 1 are the input frequency components which appear in the output 
current. The next three terms (4th, 5th, and 6th) result from the second- 
order term of the series. Two of these terms are squared and result in 
second harmonics of both of the input frequencies, as shown in Chapter 10. 
The other second-order term is the product term 2A 2 A e A m cos w c t cos <o m t. 
Using the trigonometric identity cos a cos b = £[ cos (a + b) + cos (a — b)], 
we see that this term becomes A 2 A c A m [cos (co c + co m )t + cos (co c — (o m )t]. 
In this case the frequencies (co c + co m ) and (co c — co m ) are the sum and 
difference frequencies, or the sideband frequencies, of the modulated 
wave. Thus it is seen that a second-or der nonlinearity of amplifier 
parameters causes modulatio n. 

Continuing the investigation, we see that the remaining four terms of 
Eq. 14.21 result from the third-order term of the series. If the trigono- 
metric identity cos 3 = | cos a + \ cos 3a is used, it is seen that the cubic 
cosine terms produce third harmonics of the input frequencies as well as 
contribute to the fundamental. The remaining two terms of the form 
cos 2 a cos b could be written as cos a (cos a cos b) which, in turn could be 
written as (cos a/2)[cos (a + b) + cos (a — *)]. Using the identities again, 
we could write this term as J[cos (2a + b) + cos b + cos (2a — b) — cos b]. 
When w m t is substituted for a, and o> c t for b, it is seen that sideband 
frequencies appear at twice the modulating frequency; or, in other 
words, there is distortion in the modulation components. On the other 
hand, when <o c t is substituted for a, and m m t for b, in the foregoing trigo- 
nometric term, it is seen that the second harmonic of the carrier frequency 
2a),. also has sidebands. Therefore, the cubic term of the series produces 



540 



Electronic Engineering 



second harmonic distortion of the modulation envelope, third harmonic 
distortion of the input frequencies, and sidebands of the 2co c term. From 
the preceding analysis, it should be evident that if the fourth-order term 
were included in the series, fourth harmonics of the input frequencies and 
third harmonic distortion terms of the modulation envelope would occur 
in the amplifier current, and so on. 



E 



^1 



M 



A 2 



A 3 

L 



A 3 



A 2 



A 2 



COm |3&>m 
2w m 



A 3 

L 



A 2 
A3 I A3 



w c -2a) m | <Oc |a> c + 2&>„, 



A3 



I 2o> c I 3w c 

2co c -co m 2co c + ai m 



Frequency - 



Fig. 14.7. A display of the frequency components which appear in the output current of 
a nonlinear amplifier which has two frequencies, w e and w m , applied to its input. Only 
second- and third-order nonlinearities are considered. 



The various frequency components which appear in the output current 
of the nonlinear amplifier are displayed graphically in Fig. 14.7. The 
relative magnitudes of these various components depend on the relative 
magnitudes of the first-order coefficient A x , the second-order coefficient 
A 2 , and the third-order coefficient A 3 , as shown in Fig. 14.7. 

If the amplifier has only a second-order nonlinearity and a filter were 
used to select only the carrier and the side band frequencies, the output 
of the amplifier would be (from Eq. 14.21 and the trigonometric identities) 

i = A X A C cos (o c t + /M cJ 4 m [cos (»c + W J' + cos (<o c — co m )t] (14.22) 

If we write Eq. 14.22 in the form of Eq. 14.3, 



i = AiAAcos co c t + 



AnA r 



A 1 



[cos(eo c + co m )t + cosOw,. — co m )t] 



(14.23) 

By comparison with Eq. 14.3, the modulation index of this second-order 
or "square-law" modulator is 



M = 



A 1 



(14.24) 



Amplitude Modulation and Detection 541 

A good "square-law" modulator would then have a large value of A 2 , 
or high degree of second-order nonlinearity, but no higher order of non- 
linearity. Unfortunately, this type of device may be difficult to find. 
However, examples of this type of modulated amplifier appear later in 
this chapter. 

A slight digression at this point may be in order. It should be recalled 
that in Chapter 10 the objective was to minimize the nonlinearity so that 
the output waveform would be identical to the input waveform and hence 
contain the same frequency components in the same relative magnitudes. 
The harmonics present in the output were used as an index of the degree 
of nonlinearity of the amplifier. Although the harmonics are a good 
index of the nonlinearity of an audio amplifier, which is used in the 
reproduction of sound, they are not the cause of the dischordant, un- 
pleasant sounds which come from a nonlinear amplifier. The harmonicly 
related frequencies are harmonious, and although an increased harmonic 
content will change the timbre of the sound, it will not cause the sound to 
be unpleasant. It is the modulation terms (sum and difference frequencies) 
produced by the nonlinearity which cause the dischordant, unpleasant 
sounds. The sum and difference frequencies may not be harmonious 
with the original frequencies. Thus a poor sound reproducer may sound 
acceptable for a solo performance but unacceptable for an ensemble. 

PROB. 14.5. A pentode amplifier has g m = 4000 + 200 v G fimhos. The Q 
point is V G = — 10 v. Two sinusoidal signals, each having a maximum ampli- 
tude of 5 v are applied to the input. Their frequencies are 150 Hz and 400 Hz. 
The load is 5 KQ resistive, and the amplifier is well designed. Assuming the 
load resistance to be very small in comparison with the plate resistance, determine 
the amplitude of the sum and difference frequencies and second harmonics in 
the output. What is the per cent modulation, assuming the 400 Hz frequency 
to be the carrier? Answer: fund. = 50 V peak, 2nd Har = 12.5 V peak, 
Sum = Diff = 25 V peak, % Mod = 100. 

14.4 SINGLE SIDEBAND TRANSMISSION 

In a modulated wave the sidebands carry the intelligence. Yet the total 
sideband power is usually much less than the carrier power. Also, the 
upper sideband carries the same information as the lower sideband. 
Therefore, amplitude modulation seems to be an inefficient way to trans- 
mit intelligence. To increase the efficiency of transmission, the carrier 
and o ne sideband are sometimes eliminated . This type of transmission is 
known as single sideband transmission and results in reduced power and 
bandwidth requirement for a specific transmission effectiveness. When 
single sideband transmission is used, the carrier (from a local oscillator) 
must be inserted in the receiver to recover the intelligence. As wuTbe 



542 Electronic Engineering 

shown later, the local oscillator which provides the carrier must have very 
good frequency stability to faithfully reproduce the modulating signal. 

One of the problems of single sideband transmission is the separation 
of one sideband from the other sideband and carrier. If low-frequency 
components are present in the modulating signal, some side frequencies 
are very near the carrier frequency. A commonly used circuit which 
eliminates the carrier and thus reduces the filtering requirement for single 



Modulating 
signal 



KTnJTs 



o Carrier o 




Sidebands 
(no carrier) 



Fig. 14.8. A balanced modulator circuit. 

sideband production is the balanced modulator. In a typical balanced 
modulator circuit such as is shown in Fig. 14.8, the carrier is applied in 
phase to the inputs of the two transistors while the modulating signal is 
applied in opposite phase to the two inputs. The amplifiers are operated 
class B so modulation takes place in each amplifier. The carrier is sup- 
pressed in the output because the carrier is applied "in phase" to the inputs. 
The output is tuned to the desired sideband frequency, thus essentially elim- 
inating the modulating signal. Additional filtering is required to eliminate 
the undesired sideband. 

A brief analysis will illustrate this modulation process. It will be assumed 
that the amplifier is a square law device when biased at cutoff. This is not 
strictly true, but it is a reasonable approximation when the source 
impedance is low. Then, assuming the amplifiers to be identical, we see 
that 

h = A + Av b i + A 2 v bl 2 ) 



but 



and 



= A + A-^f, 



1 2 u b2 



v b i = K c os w c t + V m cos co m t) 



V, cos m.t — V m cos a) m t) 



(14.25) 



(14.26) 



Amplitude Modulation and Detection 543 

When these values of v„ are substituted into Eq. 14.25, 
i x = A + A^V,. cos m c t + V m cos wj) + A 2 (V C cos a> c t + V m cos <oJf 
h = A + A X (V C cos m e t - V m cos coj) + A 2 (V C cos co c t - V m cos wj)* 

(14.27) 

Because of the push-pull arrangement, only the components of collector 
current which are of opposite polarity will be effective in producing an 
output. Then the effective output current is 

i e = h - h = 2A 1 V m cos m m t + 4A 2 V c V m cos w e t cos co m t (14.28) 




Fig. 14.9. The waveform of a carrier and single side frequency. 

The tuned output will eliminate the modulating frequency component, 
so the effective output voltage is 

v = B cos coj cos m m t = — [cos (co c + a) m )t + cos ((o c t — (o m t)] 

(14.29) 

where the constant B includes the tuned circuit constants in addition to 
4A 2 V c V m . It was assumed in Eq. 14.29 that the tuned output circuit 
accepted each side frequency equally well. 

The waveform of the modulation envelope is altered by the removal of 
one of the sidebands. Figure 14.9 shows the waveform of a single side 
frequency and a carrier. Note that the envelope is not sinusoidal. This 



544 Electronic Engineering 

same waveform is produced whenever two signals having nearly the same 
frequency are mixed or added. The amplitude of the combination then 
varies as the difference between their two frequencies because of the 
alternate reinforcement and cancellation of their instantaneous values. 

PROB. 14.6. Add the instantaneous values of a 1 volt 9 Hz signal to those of a 
1 volt 10 Hz to verify the waveform of Fig. 14.9. 

Both sidebands are not transmitted with the carrier suppressed because 
the carrier which must be inserted in the receiver must then maintain the 
proper phase relationship with the sidebands to have the intelligence 
properly recovered. This preciseness of oscillator control is practically 
impossible. At least the required circuit complexity would be prohibitive 
for the benefit gained. Sometimes the carrier and one sideband are trans- 
mitted to reduce the required bandwidth. This is done in modern television 
practice, where a small part of the other sideband is included to reduce the 
requirements of the filter which must discriminate between the carrier 
and the deleted sideband. 

14.5 DETECTION OF AMPLITUDE MODULATED WAVES 

The process by which the original modulating signal, or intelligence, 
is recovered in the receiving equipment is known as detection or demodu- 
lation. It was previously noted that an amplitude modulated wave normally 
consists of the carrier and the sideband frequencies only and does not 
contain the modulating frequencies. Therefore, the modulating signal, 
or frequencies, must be reproduced in the receiver to complete the trans- 
mission of the intelligence. For example, in the broadcast of entertain- 
ment, the carrier may be modulated by an orchestra. This carrier with its 
sidebands are radiated from the transmitting antenna in the form of 
electromagnetic waves. These waves in turn induce small voltages into 
the receiving antenna. These voltages are usually amplified by tuned 
amplifiers with sufficient bandwidth to include the sidebands. If the 
receiver included only linear amplifiers, the amplified carrier and side- 
bands would be fed to a loud speaker. However, this would be futile 
because the loud speaker cannot respond to the carrier or sideband 
frequencies because they are radio frequencies. Therefore, the receiver 
must include a detector. 

Since each modulating frequency is the difference between a side- 
band frequency and the carrier frequency, it seems evident that a non- 
linear device is needed to recover the modulating frequencies from the 
modulated wave. The square law device is one possible candidate for a 
detector. Again, the vacuum tube and tpansistor are essentially square-law 
devices when biased near cutoff. Thus, they may be used in this mode 



Amplitude Modulation and Detection 545 

to provide detection as illustrated in Fig. 14.10. It may be seen from 
Fig. 14.10c and 14.10*/ that the square law detector essentially eliminates 
one-half the modulated wave. In other words, the detector acts as a 
rectifier. 

It should be observed that the current pulses in the output of the detector 
vary in amplitude in accordance with the modulation envelope or modu- 
lating signal. The capacitor C (referring to Fig. 14.10) is placed in parallel 
with the load resistance for the purpose of bypassing the carrier and side- 
band frequencies so that only the recovered modulating signal will appear 
in the output. In a vacuum tube amplifier, the resistor i?j permits a 
fairly constant current to flow through R K and thus provides a fairly 
constant self-bias for the tube. This bias will fluctuate somewhat with 
signal level because the tube current is highly dependent on signal level, 
but if the bleeder current through ^ is made large in comparison with the 
average tube current, the bias may be maintained near cutoff for a wide 
range of signal levels. The cathode bypass capacitor C K should be large 
enough to bypass the detected modulating frequencies. 

We assume the detector to have only second-order nonlinearity, the 
output current i is 

i = A + A lVi + A 2 v? (14.30) 

If the input voltage is a carrier and two side frequencies which were 
produced by a sinusoidal modulating signal having the natural frequency 
co m , then from Eq. 14.3, 

MV MV 

v t = V c cos (o c t + —z cos (w c + co m )t + — — - cos (co c - cojt 

(14.31) 
Substituting this value of v t into Eq. 14.30, we have 



'o = A o + A \ 
+ A 



MV MV 

V c cos (o c t -\ — £ cos(eo (i + co m )t + — £ cos((o e — u) m )t 



MV MV 
V c cos m c t -\ cos {co c + co m )t H £ cos (w c — a> m )t 



(14.32) 

Since the bypass and d-c blocking capacitors in the output eliminate all 
components of output voltage except those which may result from the 
squared term of Eq. 14.32, the useful value of output current is 

i' B = A 2 l V c 2 cos 2 m c t + MV C 2 cos io e t[cos(fi} c + co m )f + cos(co c — co m )t] 

+ vy^W (<°° + ft) «-) r + cos ( w ° ~ m ^\ (14 - 33) 



546 



Electronic Engineering 




Mod. 
■j(— ° signal 
output 



(a) 




Fig. 14.10. (a) Vacuum tube detector; (b) transistor detector; (c) and (d) illustration 
of the detection process by the use of the transfer characteristics. 



Amplitude Modulation and Detection 



547 



r-f 
In 



4k 




l» Mr 



Mod. 
signal 
output 



(b) 




(d) 



Fig. 14.10. {Continued) 



548 Electronic Engineering 

If we expand Eq. 14.33 and retain only those products which can produce 
components which will be retained in the output, i' becomes 

'"„ = A 2 MV C 2 cos co c t cos (cd c + w m )t + A 2 MV 2 cos co c t cos (co c — co m )t 

A M 2 V 2 
+ ° cos (co c + co m )t cos (w c — cojt (14.34) 

When we replace the product terms of Eq. 14.34 with their trigonometric 
identities and retain only the terms which will be retained in the output, 
i" becomes 

.„, A 2 MV 2 ( M „ \ 

i o = — I cos u) m t + cos a> m t -\ cos 2w m t\ (14.35) 

After filtering, the output voltage is 

v = A 2 MV c 2 R L l cos co m t -\ cos 2w m t\ (14.36) 

Observe from Eq. 14.36 that a second harmonic of the modulating 
frequency appears in the output, and that the amplitude of this second 
harmonic is one-fourth M as large as the amplitude of the recovered 
modulating frequency. 

PROB. 14.7. Prove that the square law detector will recover the modulating 
signal without distortion when single sideband transmission is used. 
PROB. 14.8. If single sideband transmission were used and the carrier sup- 
pressed at the transmitter, what would be the effect on the detector output fre- 
quencies if the local oscillator which provides the carrier in the receiver were 
tuned to co. + Aco instead of m. ? 



14.6 LINEAR DETECTORS 

In reference to Fig. 14.10c and 14.1(k/, it seems that distortionless (or 
linear) detection could be accomplished by a linear rectifier. A linear 
rectifier is a device which conducts only during alternate half cycles of the 
input signal, and during the conducting half cycles the output current is 
proportional to the input voltage. The characteristics of a linear rectifier 
and the distortionless detection which may be obtained with this rectifier 
are shown in Fig. 14.11. It may be seen from the figure that the peak 
amplitude of each current pulse in the output is proportional to the 
peak amplitude of the input voltage during that particular conducting 
half cycle. Thus the peak, and therefore the average, values of the output 
current pulses follow the amplitude of the input voltage precisely during 
conducting half cycles and have the same waveform as the modulation 



Amplitude Modulation and Detection S49 

envelope. Whether the output voltage would approach the peak or average 
value of the input voltage depends on the type of filter used in the output. 
If the filter is a bypass capacitor, as previously indicated, and the internal 
resistance of the rectifier is small in comparison with the load resistance, 
the output voltage will tend to follow the peaks of the input voltage as 
explained in Chapter 3. Thus, the capacitor charges essentially to the 
peak input voltage during the conducting half cycles, but there is not time 




Fig. 14.11. The detection characteristics of a linear rectifier. 

for appreciable discharge through the high resistance load during the 
non-conducting half cycles. If the bypass capacitor is too large, the time 
constant of the discharge will be so large that the detector output will not 
be able to follow the modulation envelope when the modulation envelope 
decreases amplitude rapidly. This situation, which may occur when the 
modulating frequencies are high, is known as negative clipping and will 
be discussed in more detail later. 

As discussed in Chapter 3, the vacuum diode or semiconductor diode 
has a linear dynamic characteristic when the load resistance is large in 
comparison with the internal resistance of the diode. The voltage drop 
across the diode then becomes insignificant in comparison with the load 
voltage which then closely follows the input voltage. Thus, the properly 
designed diode detector may be a linear detector. Typical diode detector 
circuits are shown in Fig. 14.12. In the semiconductor diode circuit, 
Fig. 14.12a, R d is the diode load resistor, C f is the r-f bypass capacitor 



550 



Electronic Engineering 



r-f 

In 



I 







Ri< Rd< 



AGC 

-« — 



:Ci 



T 
I 



c f R L 



(a) 




— 




L 


r-f 




3 


In 




a 




(b) 
Fig. 14.12. (a) A typical semiconductor diode detector and AGC circuit, (b) A duo- 
diode triode tube used as a combination detector, AGC, and audio amplifier. 



and C c is the d-c blocking and load-coupling capacitor. R L represents 
the actual load on the detector. In this circuit, R L is a gain control (called 
a volume control in audio circuits). 

In addition to detection, a circuit has been added which is called the 
automatic gain control (AGC). The AGC voltage is the average value of 
the detector output voltage since R L and C x act as a filter to remove the 
modulating signal as well as the r-f from the AGC system. This AGC 
voltage is therefore proportional to the amplitude of the carrier in a 
continuous wave system and may be used to automatically control the gain 



Amplitude Modulation and Detection 551 

of one or more r-f amplifier stages. For example, if p-n-p transistors are 
used as r-f amplifiers ahead of the detector shown in Fig. 14.12a and the 
diode is connected as shown, the AGC voltage will tend to reverse bias 
the controlled amplifiers and hence reduce their gain. Thus, for small 
input signals the r-f amplifier will have high gain, but as the magnitude of 
the input signal increases, the gain of the r-f amplifier decreases. This 
effect tends to keep the detector output relatively constant and prevents 
over-driving the r-f amplifiers. Overdriving an amplifier means to cause 
its operation to extend into the saturation and cutoff regions. This type 
of operation in an r-f amplifier changes the waveform of the modulation 
envelope, thus causing severe distortion. 

The circuit of Fig. 14.126 has a few added features that are worthy 
of mention. Double tuning has been used in the input coupling circuit. 
The popular duo-diode triode tube has been used as a combination 
detector, AGC and audio amplifier circuit. Since the amplifier section 
uses cathode bias, the detector load resistor R d is returned to the cathode 
rather than to the ground in order to prevent biasing the detector plate as 
well as the amplifier grid. A Tr-type r-f filter, consisting of the resistor R f 
and the two capacitors C f , has been used to reduce the r-f in the output. 
The AGC voltage is obtained from a second diode plate which is coupled 
to the first by a capacitor. The AGC diode load resistor is R 2 and the AGC 
filter is R 1 and C t . This AGC arrangement is known as delayed AGC 
because the AGC diode is reverse-biased. This reverse bias is provided by 
the cathode bias resistor since the AGC load resistor returns to ground. 
Thus no AGC voltage will appear when the peak amplitude of the input 
signal is less than the bias on the AGC diode plate. Therefore, this type 
of AGC gives improved r-f gain for very small signals. 

Since the diode load resistor develops a d-c voltage which tends to 
reverse-bias the diode, it may seem that the diode may not conduct during 
the negative half cycles of the modulation envelope. To investigate this 
possibility, the relationship between the d-c output voltage and the r-f 
input voltage is shown in Fig. 14.13 for a typical diode detector. In this 
figure, it is assumed that the diode load resistor is bypassed for r-f so the 
d-c load voltage approaches the peak values of the r-f input voltage. 
However, the d-c load voltage is somewhat less than the peak input voltage 
because of the internal resistance of the diode and the partial discharge 
of the r-f bypass capacitor between the input voltage peaks as discussed 
in Chapter 3. Thus, the d-c load voltage decreases as the load resistance 
decreases. Simultaneously, the diode current increases, because both the 
voltage drop across the diode and the discharge between the input peaks 
increase with decreasing load resistance, as previously discussed. 

Since the axes of the plot of Fig. 14.13 are the load voltage and load 



552 



Electronic Engineering 

500 



400 



















(/) 


/* 




I st 




/° 




\° 


3 






I" 




Ih 




" 


c 

-* 1 

OJ / 




*r 1 












£/ 




\ / 


/■a-c Ic 


ad line / 






[ 




^^c 


) / 




















\ / 


-'d-c load line / 






--■?£o*o 






^g/ / 




















~~ — - 


jT~~| — 




\ / 
v./ 
















y~~-- 





300 



200 



1 q en 
100 



40 



30 



20 



10 



d-c volts developed across load 
Fig. 14.13. Rectification characteristics of a diode detector. 

current, load lines may be drawn on the rectification characteristics as 
shown. The use of the load line may be most easily explained by an 
example. 

Example 14.2. Consider a 150 Kfl d-c load resistance for the diode of Fig. 14.13. 
Assuming that the peak carrier voltage at the input of the detector is 20 v, the d-c 
voltage V Q across the diode load resistor is about 17.5 v as indicated. So long as 
the carrier is unmodulated, the d-c voltage across the load resistor will remain 
constant and the quiescent operating point on the load line will be at point Q. 
On the other hand, when the carrier is modulated the operating point must move 
up and down along the d-c load line, since the detector input voltage is continu- 
ally changing. This condition is true if the d-c load is the only load on the 
detector. When the carrier has 100% modulation, the input voltage must vary 
between and 40 v peak, so the output voltage will vary between and 36 v 
peak. Thus, the detection is essentially linear. 

As indicated in Fig. 14.12, the detector usually has part of its load isolated 
from the d-c load resistor by a blocking capacitor. Therefore, the a-c load 
resistance for the detector is usually smaller than the d-c load resistance. Con- 
sequently, an a-c load line should be drawn through the point Q as shown in 
Fig. 14.13. In this case, the detector operating point moves up and down along 
the a-c load line as the modulated signal is applied to the input. Thus, a modu- 
lation crest causes a greater increase of rectified current than would occur if the 
d-c load resistor had been the total load. Similarly, the rate of decrease of rectifier 



Amplitude Modulation and Detection 553 

current is more rapid during a trough, or negative half cycle, of the modulation 
than it would be if only the d-c load were present. Notice from Fig. 14.13 that 
this effect causes the diode current to be reduced to zero before the input voltage 
is reduced to zero. This is known as negative clipping. Notice that negative 
clipping occurs only when the percentage of modulation exceeds a certain value. 
In this example, where the a-c load resistance is one-half the value of the d-c 
load resistance, the rectifier current is reduced to zero when the input voltage is 
reduced to one-half the quiescent, or carrier, value. Therefore, negative clipping 
would occur if the percentage of modulation were to exceed 50%. 

A simple relationship exists between the ratio of a-c load resistance to 
d-c load resistance and the maximum modulation index that can occur 
without clipping. Referring to Fig. 14.13, we can see that the output 
voltage which results from the unmodulated carrier is 

V Q = I Q R de (14-37) 

Also, the maximum change in the output voltage which can occur without 
negative clipping is 

(AKVax = I Q Kc ( 14 - 38 ) 

But the maximum modulation index which can occur without negative 
clipping is the ratio of (AK) raa x to V Q . Then 

(AK)max IqRoc R<ic it a ->q\ 

M max = — = -— - V+-x» 

Vq *Q K dc *dc 

At high modulating frequencies the r-f bypass capacitor may significantly 
reduce the a-c impedance and thus can cause negative clipping, as pre- 
viously mentioned. Also, it is apparent that any load which is coupled to 
the detector through a capacitor must be large in comparison with the d-c 
load resistance to permit a high percentage of modulation without 
appreciable distortion. 

PROB. 14.9. The detector of Fig. 14.12a has R a = 50 Kfl, /?j = 1 megohm 
R L = 250 KC1, C f = 500 pf, and C x = C c = 0.1 fit. Determine the maximum 
percentage of modulation which can be accepted without causing negative 
clipping. Answer: M max = 80%. 

It should be observed that a linear detector does not provide distortion- 
less demodulation when only one sideband is present, because the modu- 
lation envelope does not have the same shape as the modulating signal 
when single sideband transmission is used. On the other hand, it may be 
proven (Prob. 14.7) that a square-law detector provides distortionless 
demodulation when single sideband transmission is used. 

The impression may have been conveyed that tube amplifiers and tran- 
sistors automatically provide square-law detection when they are biased 



554 Electronic Engineering 

near cutoff. This is not true. These amplifying types of detectors may 
provide fairly linear detection if they are properly designed for this pur- 
pose. For example, the triode tube may provide linear detection if the 
load resistance is large in comparison with the plate resistance. Also, the 
proper adjustment of driving source resistance may provide good linearity 
for a transistor detector. 

In the foregoing paragraphs, the term r-f has sometimes been used to 
mean the carrier and sidebands whereas audio has been used to indicate the 
modulation. This usage is much too specialized. In some applications, 
such as automatic control, the carrier may be low frequency of a few 
hundred Hertz whereas the modulating signal may have a period of 
several seconds. 

The effective input resistance of a detector must be obtained in some 
manner before the tuned coupling circuit can be properly designed. The 
effective input resistance may be determined if the average input power can 
be found in terms of the input voltage. The average input power over one 
cycle is 



= — ) vidO (14.40) 

2-7T J-n 



Assuming the input voltage to be an unmodulated carrier so that 
v = V c cos 0, we see that 

P = — T VJ cos OdO (14.41) 

For a biased amplifier type detector or a diode detector which does not 
have an r-f bypass capacitor, the input current / flows only on alternate 
half cycles. Then if it is assumed that during these conducting half cycles 
the input current is 

V cos 
i = ^P (14.42) 

where R t is the average input resistance during the conducting half cycle, 
the average input power is 

1 [*< 2 V c 2 cos'0 
P = — — (14.43) 



2 



77 



1 InV*\ V* 



P = — — - )=— (14.44) 

IttXIRJ 4R< 



The effective input resistance may be defined as 



R = (ItmiL = YUl = 2 R t (14.45) 



Amplitude Modulation and Detection 555 

An accurate value of R ( may be difficult to obtain because of the depend- 
ence of R { on the magnitude of the input voltage in most cases. However, 
an average value of input resistance which is obtained at an average value 
of input voltage is sufficiently accurate for good design. 

The effective input resistance of a diode detector in which the load 
resistance is large in comparison with the forward resistance of the diode 
and an adequate r-f bypass capacitor is connected in parallel with the load 
resistance may be easily found if we assume that the total input power is 
dissipated in the load resistor. Since the d-c voltage across the load 
resistor R d is approximately equal to V„ the power dissipated in the load is 

v 2 
P = L°- (14.46) 

where P is the average power in the load when the carrier is unmodulated. 
But this power is approximately the same as the r-f input power which may 
be defined as 

P = YL = Yl (14.47) 

2R t R d 
where R t , again, is the effective input resistance of the detector. Therefore, 

R ~ ** (14.48) 

2 
where R d , as before, is the d-c load resistance. When the carrier is modu- 
lated it would at first seem that the a-c load resistance of the diode instead 
of the d-c resistance R d should be used to calculate the input resistance 
because additional power is coupled through the d-c blocking capacitors 
into the R-C coupled part of the load. However, this additional power 
comes from the power in the sidebands. Therefore, the effective input 
resistance remains approximately RJ2 as long as negative clipping does 
not occur. 

14.7 FREQUENCY CONVERTERS 

In the preceding paragraphs, it was shown that both modulation and 
detection involve two basic processes. The first of these processes is the 
production of sum and difference frequencies by a nonlinear device; the 
second is the separation of the desired output components from the un- 
desired ones by some type of a'filter. In the early days of radio, the idea 
was conceived that these two processes could be used to change-or translate 
any given frequency to a different frequency. The superheterodyne re- 
ceiver is based on this principle. This receiver contains an oscillator which 
is known as a local oscillator. The output from this oscillator is mixed with 



556 Electronic Engineering 

the incoming signal and the combination is applied to a nonlinear device. 
The output of this nonlinear device is tuned to the difference between the 
oscialltor frequency and the frequency of the incoming signal. The differ- 
ence frequency is known as the intermediate frequency (i-f). This system 
has two main advantages when compared with a conventional amplifier. 
First, higher gain and better stability may be obtained from an amplifier 
at the lower intermediate frequency. Second, the intermediate frequency 
may remain constant even though a wide variety of input frequencies may 
be selected because the local oscillator frequency may be varied in such a 
manner as to produce a constant difference between the oscillator fre- 
quency and the variable input frequency. Therefore, the design of the i-f 
amplifier may be optimized whereas the design of the r-f amplifier must be a 
compromise since it is required to operate over a wide range of frequencies. 

Example 14.3. The standard broadcast superheterodyne receiver is a good 
example of the principle of frequency conversion. This receiver is required to 
tune over the 550 to 1600 KHz frequency range and it should provide essentially 
constant gain and constant bandwidth over this range. A commonly used inter- 
mediate frequency is 455 KHz. To provide this intermediate frequency, the local 
oscillator must vary from 1005 to 2055 KHz and must always remain 455 KHz 
above the selected input frequency. This maintenance of a constant difference 
frequency over the band is known as tracking. The local oscillator frequency 
could be below the input signal frequency, but then its frequency range would 
have to be 95 to 1 145 KHz. This frequency ratio is too high to cover with a single 
band. When the selected input frequency is 1000 KHz and the sideband fre- 
quencies extend from 995 to 1005 KHz, the oscillator frequency is 1455 KHz. 
Then the frequencies appearing in the output current of the nonlinear device are : 

1. The original frequencies 995-1005 KHz 

1455 KHz 

2. The sum frequencies 2450-2460 KHz 

3. The difference frequencies 450-460 KHz 

4. Harmonic frequencies 1 990 KHz and upwards 

When a tuned circuit which has a resonant frequency of 455 KHz and a bandwidth 
of 10 KHz is placed in the output, the difference frequencies, which include the 
new 455 KHz carrier with its sideband frequencies, are retained and all the other 
frequencies are rejected. 

The nonlinear device which produces the sum and difference frequencies 
is frequently called a mixer. This term is not very appropriate, however, 
because it is also used to designate linear devices in which two or more 
signals are mixed, but in which additional frequencies are not produced. 
The nonlinear device is more appropriately called a first detector or 
modulator. The combination of the first detector and the local oscillator 
is known as a frequency converter. 



Amplitude Modulation and Detection 

Antenna 



V 



557 



Loudspeaker 



r-f 
amplifier 



First 
detector 



i-f 
amplifier 



Second 
detector 



amplifier ""LI 



Local 
oscillator 



Fig. 14.14. The block diagram of a typical superheterodyne receiver. 

The block diagram of a typical superheterodyne receiver is shown in 
Fig. 14.14. The r-f amplifier shown is frequently omitted in entertainment- 
type receivers. The advantages of including this r-f amplifier are briefly 
discussed in a following paragraph. 

Some typical converter circuits are shown in Fig. 14.15. In these cir- 
cuits the oscillators are the tuned grid or base type, although other types 
could be used. The dashed lines between the input and oscillator tuned 
circuits indicate that the tuning capacitors are ganged so that single knob 
tuning may be used. In the transistor circuit, the oscillator signal is 
applied to the emitter; in the tube circuit the oscillator signal is applied 




Fig. 14.15. Typical frequency converter circuits. 



558 



Electronic Engi